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)
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 $$ \ell: \underline{x} = (-6,17,0)+t(3,-11,-65) $$