The parametric form of the equation of the line $3x-6y-2z-15=0=2x+y-2z-5$ is

The given planes are $3x-6y-2z-15=0=2x+y-2z-5$

To get a point on the line of intersection of the above two planes, substitute $z=0$

it gives $3x-6y=15,2x+y=5$

solving the above two equations, we get $P\left(3,-1,0\right)$

Suppose that the direction ratios of line are $⟨a,b,c⟩$

Hence, $3a-6b-2c=0,2a+b-2c=0$ it implies that $\frac{a}{14}=\frac{b}{2}=\frac{c}{15}$     $\left(by cross multiplication method\right)$

Therefore, the equation of the line $\frac{x-3}{14}=\frac{y+1}{2}=\frac{z}{15}$