a(x−x1)+b(y−y1)+c(z−z1)=0 ⇒−1(x+5)−2(y−19)−1(z+14)=0 P:x+2y+z=19 Any general point on the line is x=λ+2,y=−3λ+3,z=4λ+8 So, A(λ+2,−3λ+3,4λ+8) Now, this point lies on the plane ⇒(λ+2)+2(−3λ+3)+(4λ+8)=19 −λ=3⇒λ=−3 So, point of intersection A(λ+2,−3λ+3,4λ+8) A(−3+2,9+3,−12+8)≡(−1,12,−4) So position vector of A is −