I have a question asking me to find a line in the symmetrical equation that passes through $(1,1,1)$ and also parallel to the following two planes
- $\pi_1 = [x,y,z] = [1,-3,4]+r[-1,3,2]+s[2,1,2]$
- $\pi_2 = [x,y,z] = [1,2,1]+r[1,-3,2]+s[1,1,2]$
I manage to find the normal equation of these two planes:
- $\pi_1 = 4x+6y-7z=-42 $
- $\pi_2 = -8x+4z=-4 $
Then I am not sure how to proceed with this question. However, we can see that these two planes are not even parallel to each other, so how can we find a line that is parallel two both of them?
The answer key provided the solution as:
$\frac{x-1}{3}=\frac{y-1}{5}=\frac{z-1}{6}$, but I drew the line in Geogebra and I do not think the line is parallel to both planes. 
To be parallel to both planes, the direction vector $\vec v$ of the line must be orthogonal to both normal vectors, then a direction vectors is given by
$$\vec v=\vec n_1\times \vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\4&6&-7\\-8&0&-4\end{vmatrix}$$