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: $\frac{1}{2}(7,3,-3)$


I get $(3.5,1.5,-1.5)^T$ as well.
The yellow line is the intersection of the two planes, the big red dot the intersection with the given line (black).
$$
E_1: 3x - 2y + z = 6 \wedge E_2: x + y - 2z = 8
$$
Solving for $z$ gives
$$
z = 6 - 3x + 2y = -4 + x/2 + y/2 \\
10 = 7/2 x - 3/2 y \iff y = 7/3 x - 20/3
$$
E.g.choosing $x=3$ and $x=4$ then $a = (3, 1/3, 6 - 9 + 2/3)^T = (3, 1/3, -7/3)^T$ and 
$b = (4, 8/3, 6-12+16/3)^T = (4, 8/3, -2/3)^T$ are part of the intersection $E_1 \cap E_2$.
From this we generate the line
\begin{align}
g &: (3, 1/3,-7/3)^T + ((4, 8/3,-2/3)^T - (3, 1/3,-7/3)^T) s \\
  &= (3, 1/3,-7/3)^T + (1, 7/3, 5/3)^T s
\end{align}
and intersect it with the given line
$$
f: (1,1,-1)^T + (5,1,-1)^T t
$$
$g(s) = f(t)$, which gives the system:
$$
\left(
\begin{matrix}
1   & -5 \\
7/3 & -1 \\
5/3 &  1
\end{matrix}
\right)
\left(
\begin{matrix}
s \\
t
\end{matrix}
\right) =
\left(
\begin{matrix}
-2 \\
2/3 \\
4/3
\end{matrix}
\right)
$$
This has the solution $(s, t) = (1/2, 1/2)$.
Using for example $t=1/2$ with $f$, we get 
$$
P = (3.5, 1.5, -1.5)^T
$$
