P1:x+3y−z=6 P2:−6x+5y−z=7 Family of planes passing through line of intersection of P1 and P2 is given by x(1−6λ)+y(3+5λ)+z(−1−λ)−(6+7λ)=0 It passes through (2,3,‌
1
2
) ‌ So, ‌2(1−6λ)+3(3+5λ)+‌
1
2
(−1−λ)−(6+7λ)=0 ⇒2−12λ+9+15λ−‌
1
2
−‌
λ
2
−6−7λ=0 ⇒‌
9
2
−‌
9λ
2
=0⇒λ=1 Required plane is −5x+8y−2z−13=0 ‌ Or ‌