The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Y-y1=m(x-x1)
a point on the line is (x1,y1) and the slope is m
so
slope between (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)
so
points (-2,3) and (3,0)
slope is (0-3)/(3-(-2))=-3/(3+2)=-3/5
a point is (3,0) Or (-2,3)
so it could be
y-0=-3/5(x-3) or y-3=-3/5(x-(-2)) which is equal to y-3=-3/5(x+2)
that's the answer
B is answer
