Trong không gian Oxyz, cho tam giác nhọn ABC có H(2;2;1),\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaabm % aabaGaeyOeI0YaaSaaaeaacaaI4aaabaGaaG4maaaacaGG7aGaaGPa % VpaalaaabaGaaGinaaqaaiaaiodaaaGaai4oaiaaykW7daWcaaqaai % aaiIdaaeaacaaIZaaaaaGaayjkaiaawMcaaaaa!4277! K\left( { - \frac{8}{3};\,\frac{4}{3};\,\frac{8}{3}} \right)\) , O  lần lượt là hình chiếu vuông góc của A , B, C trên các cạnh BC, AC,AB . Đường thẳng d qua A và vuông góc với mặt phẳng (ABC) có phương trình là

Ta có tứ giác BOKC là tứ giác nội tiếp đường tròn suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadkeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % doeacaWGcbaacaGLcmaaaaa!3D4C! \widehat {OKB} = \widehat {OCB}\) (1)

Ta có tứ giác KDHC là tứ giác nội tiếp đường tròn suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGebGaam4saiaadIeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % doeacaWGcbaacaGLcmaaaaa!3D47! \widehat {DKH} = \widehat {OCB}\)(2)

Từ (1) và (2) suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGebGaam4saiaadIeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % dUeacaWGcbaacaGLcmaaaaa!3D4F! \widehat {DKH} = \widehat {OKB}\) .

Do đó BK là đường phân giác trong của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) và AC là đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) .

Tương tự ta chứng minh được OC là đường phân giác trong của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGlbGaam4taiaadIeaaiaawkWaaaaa!3927! \widehat {KOH}\) và AB là đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGlbGaam4taiaadIeaaiaawkWaaaaa!3927! \widehat {KOH}\).

Ta có OK = 4; OH = 3; KH = 5

Gọi I, J lần lượt là chân đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) và \(\widehat {KOH}\).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9iaadgeacaWGdbGaeyykICSaamisaiaad+eaaaa!3C95! I = AC \cap HO\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGjbGaam4taaqaaiaadMeacaWGibaaaiabg2da9maalaaabaGaam4s % aiaad+eaaeaacaWGlbGaamisaaaacqGH9aqpdaWcaaqaaiaaisdaae % aacaaI1aaaaaaa!402B! \frac{{IO}}{{IH}} = \frac{{KO}}{{KH}} = \frac{4}{5}\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGjbGaam4taaGaay51GaGaeyypa0ZaaSaaaeaacaaI0aaa % baGaaGynaaaadaWhcaqaaiaadMeacaWGibaacaGLxdcaaaa!4189! \Rightarrow \overrightarrow {IO} = \frac{4}{5}\overrightarrow {IH} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ysamaabmaabaGaeyOeI0IaaGioaiaacUdacaaMc8UaeyOeI0IaaGio % aiaacUdacaaMc8UaeyOeI0IaaGinaaGaayjkaiaawMcaaaaa!4445! \Rightarrow I\left( { - 8;\, - 8;\, - 4} \right)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOsaiabg2 % da9iaadgeacaWGcbGaeyykICSaam4saiaadIeaaaa!3C91! J = AB \cap KH\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGkbGaam4saaqaaiaadQeacaWGibaaaiabg2da9maalaaabaGaam4t % aiaadUeaaeaacaWGpbGaamisaaaacqGH9aqpdaWcaaqaaiaaisdaae % aacaaIZaaaaaaa!402B! \frac{{JK}}{{JH}} = \frac{{OK}}{{OH}} = \frac{4}{3}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGkbGaam4saaGaay51GaGaeyypa0ZaaSaaaeaacaaI0aaa % baGaaG4maaaadaWhcaqaaiaadQeacaWGibaacaGLxdcacqGHshI3ca % WGkbWaaeWaaeaacaaIXaGaaGOnaiaacUdacaaMc8UaaGinaiaacUda % caaMc8UaeyOeI0IaaGinaaGaayjkaiaawMcaaaaa!4EB2! \Rightarrow \overrightarrow {JK} = \frac{4}{3}\overrightarrow {JH} \Rightarrow J\left( {16;\,4;\, - 4} \right)\).

Đường thẳng IK qua I nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaam4saaGaay51GaGaeyypa0ZaaeWaaeaadaWcaaqaaiaaigda % caaI2aaabaGaaG4maaaacaGG7aGaaGPaVpaalaaabaGaaGOmaiaaiI % daaeaacaaIZaaaaiaacUdacaaMc8+aaSaaaeaacaaIYaGaaGimaaqa % aiaaiodaaaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI0aaaba % GaaG4maaaadaqadaqaaiaaisdacaGG7aGaaGPaVlaaiEdacaGG7aGa % aGPaVlaaiwdaaiaawIcacaGLPaaaaaa!522B! \overrightarrow {IK} = \left( {\frac{{16}}{3};\,\frac{{28}}{3};\,\frac{{20}}{3}} \right) = \frac{4}{3}\left( {4;\,7;\,5} \right)\) làm vec tơ chỉ phương có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGjbGaam4saaGaayjkaiaawMcaaiaacQdadaGabaabaeqabaGaamiE % aiabg2da9iabgkHiTiaaiIdacqGHRaWkcaaI0aGaamiDaaqaaiaadM % hacqGH9aqpcqGHsislcaaI4aGaey4kaSIaaG4naiaadshaaeaacaWG % 6bGaeyypa0JaeyOeI0IaaGinaiabgUcaRiaaiwdacaWG0baaaiaawU % haaaaa!4DDF! \left( {IK} \right):\left\{ \begin{array}{l} x = - 8 + 4t\\ y = - 8 + 7t\\ z = - 4 + 5t \end{array} \right.\).

Đường thẳng OJ qua O nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGpbGaamOsaaGaay51GaGaeyypa0ZaaeWaaeaacaaIXaGaaGOnaiaa % cUdacaaMc8UaaGinaiaacUdacaaMc8UaeyOeI0IaaGinaaGaayjkai % aawMcaaiabg2da9iaaisdadaqadaqaaiaaisdacaGG7aGaaGPaVlaa % igdacaGG7aGaaGPaVlabgkHiTiaaigdaaiaawIcacaGLPaaaaaa!4F54! \overrightarrow {OJ} = \left( {16;\,4;\, - 4} \right) = 4\left( {4;\,1;\, - 1} \right)\) làm vec tơ chỉ phương có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGpbGaamOsaaGaayjkaiaawMcaaiaacQdadaGabaabaeqabaGaamiE % aiabg2da9iaaisdaceWG0bGbauaaaeaacaWG5bGaeyypa0JabmiDay % aafaaabaGaamOEaiabg2da9iabgkHiTiqadshagaqbaaaacaGL7baa % aaa!45C6! \left( {OJ} \right):\left\{ \begin{array}{l} x = 4t'\\ y = t'\\ z = - t' \end{array} \right.\).

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iaadMeacaWGlbGaeyykICSaam4taiaadQeaaaa!3C9F! A = IK \cap OJ\), giải hệ ta tìm được A(-4 ; -1; 1).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamyqaaGaay51GaGaeyypa0ZaaeWaaeaacaaI0aGaai4oaiaa % ykW7caaI3aGaai4oaiaaykW7caaI1aaacaGLOaGaayzkaaaaaa!429D! \overrightarrow {IA} = \left( {4;\,7;\,5} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamOsaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGaaGinaiaa % cUdacaaMc8UaaGymaiaaikdacaGG7aGaaGPaVlaaicdaaiaawIcaca % GLPaaaaaa!4413! \overrightarrow {IJ} = \left( {24;\,12;\,0} \right)\) , ta tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaada % WhcaqaaiaadMeacaWGbbaacaGLxdcacaGGSaWaa8HaaeaacaWGjbGa % amOsaaGaay51GaaacaGLBbGaayzxaaGaeyypa0ZaaeWaaeaacqGHsi % slcaaI2aGaaGimaiaacUdacaaIXaGaaGOmaiaaicdacaGG7aGaeyOe % I0IaaGymaiaaikdacaaIWaaacaGLOaGaayzkaaGaeyypa0JaeyOeI0 % IaaGOnaiaaicdadaqadaqaaiaaigdacaGG7aGaaGPaVlabgkHiTiaa % ikdacaGG7aGaaGPaVlaaikdaaiaawIcacaGLPaaaaaa!579C! \left[ {\overrightarrow {IA} ,\overrightarrow {IJ} } \right] = \left( { - 60;120; - 120} \right) = - 60\left( {1;\, - 2;\,2} \right)\).

Khi đó đường thẳng đi qua A và vuông góc với mặt phẳng (ABC) có véc tơ chỉ phương \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaigdacaGG7aGaeyOeI0Ia % aGOmaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3FCF! \overrightarrow u = \left( {1; - 2;2} \right)\) nên có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGinaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaey4kaSIaaGymaaqaaiabgkHiTiaaikdaaaGaeyypa0ZaaSaaae % aacaWG6bGaeyOeI0IaaGymaaqaaiaaikdaaaaaaa!432F! \frac{{x + 4}}{1} = \frac{{y + 1}}{{ - 2}} = \frac{{z - 1}}{2}\).