Find 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

Let general point of line be A(2λ+1,4λ+3,3λ+2). Let this point lies at the same distance as  the point p (3,8,2) from the plane 3x+2y−2z+15=0 Therefore, 3.3+2.8−2.2+1517=3(2λ+1)+2(4λ+3)−2(3λ+2)+1517⇒36=8λ+20⇒λ=2 Therefore, A is (5,11,8)PA=(5−3)2+(11−8)2+(8−2)2=4+9+36=7