The position vectors of vertices of Δ ABC are A (2 i - j + k ), B ( i -3 j -5 k ) and C (3 i -4 j -4 k ). Find all the angles of Δ ABC.

$$\begin{aligned} & \cos \mathrm{A}=\frac{\overrightarrow{\mathrm{A} \mathrm{B}} \cdot \overrightarrow{\mathrm{A} \mathrm{C}}}{|\overrightarrow{\mathrm{A} \mathrm{B}}||\overrightarrow{\mathrm{A} \mathrm{C}}|}=\frac{(-\hat{\imath}-2 \hat{\jmath}-6 \hat{\mathrm{k}}) \cdot(\hat{\imath}-3 \hat{\jmath}-5 \hat{\mathrm{k}})}{\sqrt{41} \sqrt{35}}=\frac{35}{\sqrt{41} \sqrt{35}}=\frac{\sqrt{35}}{\sqrt{41}} \\ & \mathrm{A}=\cos ^{-1}\left(\frac{\sqrt{35}}{\sqrt{41}}\right) \\ & \cos \mathrm{B}=\frac{\dfrac{\overrightarrow{\mathrm{B} \mathrm{A}} \cdot \overrightarrow{\mathrm{B} \mathrm{C}}}{|\overrightarrow{\mathrm{B} \mathrm{A}}||\overrightarrow{\mathrm{B} \mathrm{C}}|}}{(\hat{\imath})}=\frac{(\hat{\imath}+2 \hat{\jmath}+6 \hat{\mathrm{k}}) \cdot(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})}{\sqrt{41} \sqrt{6}}=\frac{6}{\sqrt{41} \sqrt{6}}=\frac{\sqrt{6}}{\sqrt{41}} \\ & \mathrm{B}=\cos ^{-1}\left(\frac{\sqrt{6}}{\sqrt{41}}\right) \\ & \cos \mathrm{C}=\frac{\dfrac{\overrightarrow{\mathrm{C} \mathrm{B}} \cdot \overrightarrow{\mathrm{C} \mathrm{A}}}{|\overrightarrow{\mathrm{C} \mathrm{B}}||\overrightarrow{\mathrm{C} \mathrm{A}}|}=\dfrac{(-2 \hat{\imath}+\hat{\jmath}-\hat{\mathrm{k}}) \cdot(-\hat{\imath}+3 \hat{\jmath}+5 \hat{\mathrm{k}})}{|\overrightarrow{\mathrm{C} \mathrm{B}}||\overrightarrow{\mathrm{C} \mathrm{A}}|}=0}{\cos \mathrm{C}=0 \quad \Rightarrow \mathrm{C}=\dfrac{\pi}{2}} \end{aligned} $$