Intersection between two planes and a line? What is the coordinates of the point where the planes: $3x-2y+z-6=0$ and $x+y-2z-8=0$ and the line: $(x, y, z) = (1, 1, -1) + t(5, 1, -1)$ intersects with eachother?
I've tried letting the line where the two planes intersect eachother be equal to the given line, this results in no solutions.
I have tried inserting the lines x, y and z values into the planes equations, this too, results in no solutions.
According to the answer sheet the correct solution is: $\frac{1}{2}(7,3,-3)$

From the line equation you know that $x$ (as well as $y$) is a function of $z$:
when $z = -1-t$, then $x = 1+5t$ and hence $x = -4-5z$.
This gives you the third equation you need:
\begin{align}
3x-2y+\phantom{1}z-6&=0\\
\phantom{1}x-\phantom{1}y-2z-8&=0\\
\phantom{1}x+0y+5z+4&=0.
\end{align}
