The values of $$a$$, for which the points $$A, B, C$$ with position vectors $$2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k}$$ and $$a\hat{i} - 3\hat{j} + \hat{k}$$ respectively are the vertices of a right-angled triangle with $$C = \dfrac{\pi}{2}$$ are

To have a right angle at vertex $$C$$, the two sides meeting at $$C$$ must be perpendicular.
Hence the vectors $$\overrightarrow{CA}$$ and $$\overrightarrow{CB}$$ should satisfy $$\overrightarrow{CA}\cdot\overrightarrow{CB}=0$$.

Position vectors of the points are
$$\vec{A}=2\hat{i}-\hat{j}+\hat{k},\qquad \vec{B}= \hat{i}-3\hat{j}-5\hat{k},\qquad \vec{C}=a\hat{i}-3\hat{j}+\hat{k}.$$

Compute the two side vectors beginning at $$C$$:
$$\overrightarrow{CA}= \vec{A}-\vec{C}= (2-a)\hat{i}+(-1+3)\hat{j}+(1-1)\hat{k}=(2-a)\hat{i}+2\hat{j},$$
$$\overrightarrow{CB}= \vec{B}-\vec{C}= (1-a)\hat{i}+(-3+3)\hat{j}+(-5-1)\hat{k}=(1-a)\hat{i}-6\hat{k}.$$

Impose the perpendicularity condition:
$$\overrightarrow{CA}\cdot\overrightarrow{CB}=0$$
$$\Longrightarrow\ (2-a)(1-a)+2\cdot0+0\cdot(-6)=0$$
$$\Longrightarrow\ (2-a)(1-a)=0.$$

This product is zero when
$$2-a=0\quad\text{or}\quad 1-a=0,$$
giving $$a=2$$ or $$a=1.$$

Therefore, the required values of $$a$$ are 2 and 1.

Option A which is: 2 and 1