The length of the projection of the line segment joining the points  −1,−2,2  and  1,0,2on the plane  x+3y−5z−2=0 is equal to

Find the foot of the perpendiculars  P,Q  of the points A−1,−2,2  and  B1,0,2 on the plane  x+3y−5z−2=0 and then find the distance between those two points.The length PQ is projection of   ABon the planeIt implies PQ=ABsinθ, where θ is angle between the line AB¯ and the normal vector to the plane.Hence, PQ=22+221−cos2θ=81−21+23+0−54+4+01+9+25=81−8835=8×1−835=8×2735=6635                Therefore, the projection is 6635