Equation of a line that's parallel to the intersection of two planes 
Write the equation of the line that passes through point $(-6,17,0)$
and is parallel to the intersection line of the planes: $$4x+7y-z=0$$
$$7x-4y+z=3$$

my approach:
first find the intersection line of the two planes by solving a system of equations:
$$x = 7/11 - \frac{1}{3}z$$
$$y = -4/11 + \frac{1}{3}z$$
$$z - free$$
next, find two points on the line by choosing two arbitrary z values:
$$z=-\frac{12}{11}: A=(11,-8,-12)$$
$$z=-\frac{-144}{11}: B=(55,-52,-144)$$
now find direction of vector $\vec{AB}$:
$$\vec{AB} = (44,-44,-132) = (1,-1,3)$$
therefore, the equation of the required line: $ \ell: \underline{x} = (-6,17,0)+t(1,-1,3) $.
However this answer was marked as incorrect (I don't have the final correct answer since its an online form)
 A: You made a mistake when finding your line equation, it should have been
$$x = \frac{21}{65} - \frac{3}{65}z$$
$$y = -\frac{12}{65} + \frac{11}{65}z$$
$$z - free$$
If we follow you thoughts
next, find two points on the line by choosing two arbitrary z values:
$$z=0: A=\left(\frac{21}{65},-\frac{12}{65},0\right)$$
$$z=1: A=\left(\frac{18}{65},-\frac{1}{65},1\right)$$
now find direction of vector $\vec{AB}$:
$$\vec{AB} = \left(-\frac{3}{65},\frac{11}{65},1\right)$$
Here is a simplier one we could use
$$\vec{v} = (-3,11,65)$$
therefore, the equation of the required line:
$$ \ell: \underline{x} = (-6,17,0)+t(-3,11,65) $$

An other way of doing this with cross product. The line will be perpendicular to both normal vectors of the planes.
$$\vec{n_1} = (4, 7, -1)$$
$$\vec{n_2} = (7, -4, 1)$$
$$\vec{n_1}\times \vec{n_2} = (3, -11, -65)$$
And we have a similar answer
$$ \ell: \underline{x} = (-6,17,0)+t(3,-11,-65) $$
A: A side remark. The choice of $z-$free is awkward. With $y-$free
$$M=(x,y,z)\in \text{intersection line of the planes}\iff $$
$$\begin{cases} 4x+7y-z=0  \\ 7x-4y+z=3  \end{cases}\iff \begin{cases} z+7x-4y=3 \\ -z+4x+7y=0   \end{cases}$$
$$\iff \begin{cases} z+7y-4y=3  \\ 11x+3y=0\end{cases}\iff M\in (0,0,3)+\Bbb R(-3,11,65)$$
So, $$l=(-6,17,0)+\Bbb R(-3,11,65)$$
