Show that the points A, B and C with position vectors,

Position vectors of points A, B, and C are respectively given as:

$\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$

$\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$

$\therefore \overrightarrow{\mathrm{AB}}=\vec{b}-\vec{a}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k}$

$\overrightarrow{\mathrm{BC}}=\vec{c}-\vec{b}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}$

$\overrightarrow{\mathrm{CA}}=\vec{a}-\vec{c}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k}$

$\therefore|\overrightarrow{\mathrm{AB}}|^{2}=(-1)^{2}+3^{2}+5^{2}=1+9+25=35$

$|\overrightarrow{B C}|^{2}=(-1)^{2}+(-2)^{2}+(-6)^{2}=1+4+36=41$

$|\overrightarrow{C A}|^{2}=2^{2}+(-1)^{2}+1^{2}=4+1+1=6$

$|\overrightarrow{\mathrm{AB}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}=35+6=41=|\overrightarrow{\mathrm{BC}}|^{2}$

Hence, ABC is a right-angled triangle.