How to find the distance between a plane and a line in vectors ?

Find the distance between the line l: r = -i + 3k + u(i+j+2k) and the plane r.(i+j-k) = 3



Can someone help me pls ?How to find the distance between a plane and a line in vectors ?First note that the distance between a line and a plane is only defined if the line is parallel to the plane. Now line l is in the direction (i+j+2k) whereas the plane is normal to (i+j-k) so we need to check if these two vectors are perpendicular to each other. In fact (i+j+2k) .(i+j-k) = 1+1-2=0 so line l is indeed parallel to the plane.
Given that P = -i + 3k is a point on line l, we can measure the distance between P and the plane by passing a line through P which is perpendicular to the plane, finding where this line intersects the plane (Q, say) and measuring the distance PQ. Now, the line which goes through P and is perpendicular to the plane is r= P + v (i+j-k). Q will be the point for which
[ P + v(i+j-k) ] .(i+j-k) =3, which simplifies to:
(-i+3k) . (i+j-k) + 3v = 3, or
-4 + 3v = 3, so v=7/3. Q is therefore
Q = P + (7/3) (i+j-k) and therefore
Q-P = (7/3) (i+j-k)
The distance between line l and the plane is then the absolute value of Q-P which is 7sqrt(3)/3