Intersection between two three-dimensional planes The intersection of the planes defined by
$x \bullet \begin{pmatrix} 8 \\ 1 \\ -12 \end{pmatrix} = 35$
and
$x \bullet \begin{pmatrix} 6 \\ 7 \\ -9 \end{pmatrix} = 70$
is a line. Find an equation of this line.
I've been attempting this problem for hours and cannot solve it. I would love a few comprehensive hints so I can solve this problem. 
 A: Let $x=(a,b,c)$. Then we can rewrite your equations as
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
6a+7b-9c&=70
\end{align}$$
Now we want to solve those simultaneous equations. Subtracting $3/4$ of the first equation from the second we get
$$\begin{align}
8a+1b-12c&=35 \\[2 ex]
0a+\frac{25}4b-0c&=\frac{175}4
\end{align}$$
Dividing the first equation by $8$ and multiplying the second equation by $\frac 4{25}$ we get
$$\begin{align}
1a+\frac 18b-\frac 32c&=\frac{35}8 \\[2 ex]
0a+1b-0c&=7
\end{align}$$
Subracting $\frac 18$ of the second equation from the first we get
$$\begin{align}
1a+0b-\frac 32c&=\frac{7}2 \\[2 ex]
0a+1b-0c&=7
\end{align}$$
Our final solution is then
$$a=\frac 72+\frac 32c,\quad b=7$$
with no restriction on $c$. If we let our parameter be $t$ and let that be $c$, we see that our line has the points
$$\begin{align}
x&=\left(\frac 72+\frac 32t, 7, t\right) \\[2 ex]
 &=\left(\frac 72,7,0\right)+t\cdot\left(\frac 32,0,1\right)
\end{align}$$
If you understand matrices, you could have gotten here faster by putting the coefficients of the two equations into a $2\times 4$ matrix and reducing it to row-reduced echelon form.
If you dislike fractions, you could choose another base point on the line where $t=1$ and multiply the direction vector by $2$, getting
$$\begin{align}
x&=(5+3t, 7, 1+2t) \\[2 ex]
 &=(5,7,1)+t\cdot (3,0,2)
\end{align}$$
A: Some hints:
To find the line of intersection, you need to find the direction vector $\mathbf{V}$ of the line, and a point $\mathbf{P}$ that lies on it.
The direction vector $\mathbf{V}$ must be perpendicular to the normal vectors of the two given planes. You can construct such a vector by using a vector cross product.
To get a point $\mathbf{P}$ that lies on the line of intersection, you can intersect the two given planes with some third plane. You can choose this third plane more-or-less at random, and, unless you're very unlucky, it will work. So, to make your life easier, you may as well choose some very simple plane, like the plane $z=0$.
