#### Answer

$(x,y,z) = (12,6,0)+n(-2,3,1)$

#### Work Step by Step

$x+2z = 12$
$y-3z = 6$
The normal vectors to the planes are $a = (1,0,2)$ and $b = (0,1,-3)$ respectively.
We can find $a \times b$:
$a \times b = (0-2, 0-(-3), 1-0) = (-2, 3, 1)$
Then $(-2,3,1)$ is the direction vector of the line of intersection.
We can let $z=0$ to find a point on the line of intersection:
$x+2(0) = 12$
$x = 12$
$y-3(0) = 6$
$y = 6$
Therefore, the point $(12,6,0)$ is on the line of intersection.
The can write the equation for the line of intersection:
$(x,y,z) = (12,6,0)+n(-2,3,1)$