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Yeah! That's exactly what I did.

First I broke $ \vec{p}(t) $ and $ \vec{q}(t) $ into the other form to get their direction vectors:

$ \vec{p}(t) = \langle 3,1,4 \rangle + \langle 2,-1,1 \rangle t$ and $ \vec{q}(t) = \langle -2,0,2 \rangle + \langle 3,2,1 \rangle t $

Then I took the cross product of $ \vec{p}(t) \times \vec{q}(t) $:

$ \vec{I}\vec{J} \vec{K} $

2 -1 1

3 2 1

$ \vec{I} (-1-2)- \vec{J}(2-3)+ \vec{K}(4+3)$

$ \langle -3,1,7 \rangle $

Then plugging this for $ \langle a,b,c \rangle $ and the intersection point for $ \langle x_0,y_0,z_0 \rangle $ into the equation:

$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $

$ -3(x-1)+1(y-2)+7(z-3)=0$

And that simplifies to:

$ -3x+y+7z= 20$