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 $C = π/2$, are

2 and 1

We have,

$\vec{AB}=-\hat i+2\hat j-6\hat k,\vec{BC}=(a-1)\hat i+6\hat k$

and $\vec{CA}=(2− a)\hat i +2\hat j$

It is given that ΔABC is right angled at C.

$∴\vec{CB}.\vec{CA}=0$

$⇒\left((a-1)\hat i+6\hat k\right).\left((2− a)\hat i +2\hat j\right)=0$

$⇒(1-a) (2-a)=0⇒a=1,2$