Find point of line in the plane The line is $$\{u \in \mathbb R^3: (-\frac{1}{3}-2z, \frac 23 +2 z, z), z \in \mathbb R\}$$,and is the result of the equation $$Au=b$$, 
where A is $$\begin{bmatrix}1&2&-2\\2&1&2\\1&5&-8\\ \end{bmatrix}$$ and b is $$\begin{bmatrix}1\\0\\3\\ \end{bmatrix}$$
The plane is 1
$$P = {a(1,0,0) + c(0,1,0), a,c \in \mathbb R}$$
I was able to get there by noticing that the third number had to be 0.
In the how to of the exercise there is another method which I didn't understand. It is
$$A\left(a\begin{bmatrix}1\\0\\0\\ \end{bmatrix} + c\begin{bmatrix}0\\1\\0\\\end{bmatrix}\right) = a\begin{bmatrix}1\\2\\1\\ \end{bmatrix} + c \begin{bmatrix}2\\1\\5\\ \end{bmatrix} = \begin{bmatrix}1&2\\2&1\\1&5\\\end{bmatrix} \begin{bmatrix}a\\c\\\end{bmatrix}$$ and that the result of the equation
$$\begin{bmatrix}1&2\\2&1\\1&5\\\end{bmatrix} \begin{bmatrix}a\\c\\\end{bmatrix}=\begin{bmatrix}1\\0\\3\\ \end{bmatrix}$$ is a=$-\frac 13 ,c= \frac 23$, and that putting those numbers in the equation for u you get that the third number is z=0
I don't understand what he is doing. I understand the matrix multiplication going on, I just don't understand why he did the final equation, or what it means.
 A: Any point in the plane is given by $[x,y,z]^T=a[1,0,0]^T+c[0,1,0]^T$, because it is a linear combination of two non-collinear vectors in that plane. Now the intersection between the line and the plane obeys both equations, so what they use is to plug in the description above into the the equation of the line. $$\begin{align}A\left(\begin{bmatrix}x\\y\\z\\ \end{bmatrix}\right)&=b\\A\left(a\begin{bmatrix}1\\0\\0\\ \end{bmatrix} + c\begin{bmatrix}0\\1\\0\\\end{bmatrix}\right)&=\begin{bmatrix}1\\0\\3\\ \end{bmatrix}\end{align}$$
We know that the matrix-vector multiplication is distributive, so after some simple manipulation, the equation can be written as 
$$\begin{bmatrix}1&2\\2&1\\1&5\\\end{bmatrix} \begin{bmatrix}a\\c\\\end{bmatrix}=\begin{bmatrix}1\\0\\3\\ \end{bmatrix}$$
This is still the equation that the point $[x,y,z]^T$ in the plane (that's why you have the $a$ and $c$ coefficients) is on the given line. What is left to do is to solve this equation, to see which particular point in the plane is also on the line. In the general case, depending on $b$, you might have an infinite number or solutions (line is in the plane), one solution (line intersects the plane), or no solution (line is parallel to the plane).
