Find an equation for the plane that contains the line v = (−1, 1, 2) + t(7, 6, 2) and is perpendicular to the plane 8x + 5y − 7z + 8 = 0.

The required plane Π contains the line 
L: (-1,1,2)+t(7,6,2)
means that Π is perpendicular to the direction vector of the line L, namely
vl=<7,6,2>

It is also required that &Pi; be perpendicular to the plane
&Pi; 1 : 5y-7z+8=0
means that &Pi; is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.

Thus the normal vector of the required plane, &Pi; can be obtained by the cross product of vl and vp, or vl x vp:
 i  j  k
7 6  2
0 5 -7
=<-42-10, 0+49, 35-0>
=<-52, 49, 35> 
which is the normal vector of &Pi;

Since &Pi; has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is
&Pi; :  -52(x-(-1))+49(y-1)+35(z-2)=0
=>
&Pi; :  -52x+49y+35z = 171

Check that normal vector of plane is orthogonal to line direction vector
<-52,49,35>.<7,6,2>
=-364+294+70
=0   ok

Ibelieve the answer is b. y=0 would make the line horizontal across the x.