Write the parametric equation of the revolution surface generated by the line when it rotates around the axis $Oz$. Write the parametric equation of the revolution surface generated by the line whose equation is $2x - y + z = 1, x + y - 3z = 2$ when it rotates around the axis $Oz$.

If we let $z=t$ and we add the two equations together we get that $$x = 1 + \frac{2}{3}t \quad y = 1+\frac{7}{3}t$$
Then we multiply the vector by the rotation matix
$$\left(\begin{array}{ccc}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{array}\right)
\left(\begin{array}{c}
1 + \frac{2}{3}t \\
1 + \frac{7}{3}t \\
t
\end{array}\right) = 
\left(\begin{array}{c}
\cos\theta\left(1+\frac{2t}{3}\right)-\sin\theta\left(1+\frac{7t}{3}\right) \\
\cos\theta\left(1+\frac{7t}{3}\right)+\sin\theta\left(1+\frac{2t}{3}\right) \\ t
\end{array}\right)
$$
I am unsure how to carry on from here.
 A: The line has equation given by the system
$ 
\left\{
\begin{array}{l}
 2x - y + z = 1\\
x + y - 3z = 2\\
\end{array}
\right.
$
The planes have normal vectors $(2,-1,1)$ and $(1,1,-3)$
The intersecting line is parallel to their cross product
$(2,-1,1)\times(1,1,-3)=(2,7,3)$
and passes through a point common to the two planes which can be found adding the two equations $3x-2z=3\to x=\frac{3+2z}{3}$ so for $z=0$ we have $x=1$ and plugging these values in one of the two planes get $y=2-x+3z=1$
So the line passes through $1,1,0$ and is parallel to $(2,7,3)$
Its parametric equation is $(x,y,z)=(1,1,0)+t(2,7,3)$
$ 
\left\{
\begin{array}{l}
 x=1+2t\\
y=1+7t\\
z=3t\\
\end{array}
\right.
$
Any point of the line rotates and generates a circle whose parametric equation is
$ 
\left\{
\begin{array}{l}
x=\sqrt{(1+2t)^2+(1+7t)^2} \cos\theta\\
y=\sqrt{(1+2t)^2+(1+7t)^2} \sin\theta\\
\end{array}
\right.
$
at a $z$ value of $z=3t$
so the parametric equation of the surface of revolution is
$ 
\left\{
\begin{array}{l}
x=\sqrt{2 + 18 t + 53 t^2} \cos\theta\\
y=\sqrt{2 + 18 t + 53 t^2} \sin\theta\\
z=3t\\
\end{array}
\right.
$
for $t\in\mathbb{R};\;\theta\in[0,2\pi)$ 
