#### Answer

$(x,y,z) = (12,10,0)+n(-3,5,1)$

#### Work Step by Step

$x+3z = 12$
$y-5z = 10$
The normal vectors to the planes are $a = (1,0,3)$ and $b = (0,1,-5)$ respectively.
We can find $a \times b$:
$a \times b = (0-3, 0-(-5), 1-0) = (-3, 5, 1)$
Then $(-3,5,1)$ is the direction vector of the line of intersection.
We can let $z=0$ to find a point on the line of intersection:
$x+3(0) = 12$
$x = 12$
$y-5(0) = 10$
$y = 10$
Therefore, the point $(12,10,0)$ is on the line of intersection.
The can write the equation for the line of intersection:
$(x,y,z) = (12,10,0)+n(-3,5,1)$