Given points are A(k,1,−1),B(2k,0,2) and C(2+2k,k,1) Let r1=length of line
AB=√(2k−k)2+(0−1)2+(2+1)2=√k2+10
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
l1=
k
√k2+10
,m1=
−1
√k2+10
,n1=
3
√k2+10
and
l2=
2
√k2+5
,m2=
k
√k2+5
,n2=
−1
√k2+5
So,l1l2+m1m2+n1n2=0
⇒
2k
√k2+10√k2+5
−
k
√k2+10√k2+5
−
3
√k2+10√k2+5
=0
⇒2k−k−3=0 ⇒k=3 For k=3, AB is perpendicular to BC.