The distance of the point  P(3,8,2) from the line 12(x−1)=14(y−3)=13(z−2) measured parallel to the plane 3x+2y−2z+15=0 is

Let A=(2λ+1,4λ+3,3λ+2)Let this point lie on the line such that AP is parallel to the plane AP¯ is perpendicular to normal to the plane.⇒AP→⊥(3i^+2j^−2k^)⇒3.(2λ−2)+2(4λ−5)−2(3λ)=0⇒λ=2therefore, A is (5,11,8), P = (3, 8, 2) , AP = 7