Q92: A vector perpendicular to the plane is:
[i... j.. k]
[1 -2 -2]
[3..7..-6]
=i(12+14)-j(0)+k(13)
=26i+13k=13(2i+k)
So a vector perpendicular to the plane is 2i+0j+k.
The plane has equation r.(2i+k)=(0i+0j+ok).(2i+0j+k).
Taking r=(xi+yj+zk) we have 2x+z=0.
R1 can be written as r=i(3+s)+j(3-s)+k(-1-2s).
'2x+z'=2(3+s)-1-2s=6+2s-1-2s=5, which does not satisfy 2x+z=0 so the line does not lie in the plane.
R1 is parallel to the plane so for any line you draw through R1 and the plane, the distance is the same.
For this perpendicular line through O we have: r=i(2p)+j(0)+k(p), where p is our variable.
We want the intersection of: r=i(3+s)+j(3-s)+k(-1-2s) and r=i(2p)+j(0)+k(p). Clearly 3-s=0 so s=3 and the point of intersection is (6,0,-7)
The distance is that between the origin and (6,0,-7)=sqrt[36+49]=sqrt[85]
R2 can be written as r=i(4+3t)+j(5+7t)+k(-8-6t).
'2x+z'=2(4+3t)+(-8-6t)=2(4+3t)-2(4+3t)=0, which does satisfy the equation of the plane so R2 lies in the plane.