The parametric equations are
$$x = 2 + 3t$$ $$y = 3-t$$ $$z= 1+t$$
Let $\alpha: A_1x + B_1y + C_1z + D_1 = 0$ and $\beta: A_2x + B_2y + C_2z + D_2 = 0$ be the equations of two intersecting planes.
To make it clear, I'm asked to go from to $(x,y,z) \to \alpha$ and $\beta$.
It can be seen that the vector lying on the of intersection of the planes has coordinates $(3,-1,1)$. Since this vector is perpendicular to both normal vectors to the planes we have that
$$B_1C_2-B_2C_1=3$$ $$C_1A_2 - C_2A_1 =-1$$ $$A_1B_2- A_2B_1 = 1$$
And here I'm stuck. I don't this is the path, since I got 3 equations in 6 unknowns. The other thing I was thinking was to plug the parametric equations in the equations of the planes, in order to get something in terms of $t$ but it doesn't look that helpful either.
The parametric equations of the line $\;D\;$ are
$x=2+3t$
$y=3-t$
$z=1+t$.
from the second, we get $\;\;t=3-y\;\;$ which we replace in the others to obtain the planes equations
$$P_1\;:\;x+3y=11$$ and $$P_2\;:\;y+z=4.$$
we check that $D\in P_1$ and $D\in P_2$.