Question

Find symmetric equations for the line of intersection of the planes. z = 4x − y − 13, z = 6x + 5y − 13

Answer 1

The intersection line of two planes is the cross product of the normal vectors of the two planes.

p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>

The direction vector of the intersection line is the cross product of the two normals, 
vl=
 i   j   k
4 -1 -1
6  5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>

To find the equation of the line, we need to find a point on the intersection line.
Equate z:  4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)

Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:
$\frac{x-0}{3}=\frac{y-0}{-1}=\frac{z-(-13)}{13}$
=>
$\frac{x}{3}=\frac{-y}{1}=\frac{z+13}{13}$