BITSAT 2017 Paper

Given points are A(k,1,−1),B(2k,0,2) and C(2+2k,k,1)
Let r1=length of line
and r2=length of lineBC=√(2)2+k2+(−1)2
=√k2+5
Now, let l1,m1,n1 be direction-cosines of line AB and l2,m2,n2 be the direction cosines of BC.
Since AB is perpendicular to BC
∴l1l2+m1m2+n1n2=0
Now and So,l1l2+m1m2+n1n2=0
⇒2k−k−3=0
⇒k=3
For k=3, AB is perpendicular to BC.