Find the shortest distance between the point $Q(11, 2, -1)$ and the line of intersection created by the planes
x .$ \left(\begin{array}{cc} 1\\ -1\\ 3\\ \end{array}\right) = 0 ~$ and x $= a \left(\begin{array}{cc} 2\\ 1\\ 2\\ \end{array}\right)~+ $ $~b \left(\begin{array}{cc} 3\\ 1\\ -3\\ \end{array}\right) $
To get the direction of the line I took $(1, -1, 3)\times(v_1 \times v_2)$ where $v_1$ and $v_2$ are the directional vectors of the second plane.
$ \left(\begin{array}{cc} 1\\ -1\\ 3\\ \end{array}\right) \times$ $ \left(\begin{array}{cc} -5\\ 12\\ -1\\ \end{array}\right) =$ $ \left(\begin{array}{cc} -37\\ -16\\ 7\\ \end{array}\right) $
and the planes go through the origin, so my line is just
$\lambda \left(\begin{array}{cc} -37\\ -16\\ 7\\ \end{array}\right) $
If I imagine a triangle created by the line going from the line of the intersection to the point Q as being the hypotenuse, then if I project Q onto the direction $(-37, -16, 7)$, I get the magnitude of the base of the triangle
$\frac{446}{\sqrt{1674}}$
and all I need to do is use Pythagoras to find the last side of my triangle which will be the perpendicular distance from the line to my point. Therefore I get:
Shortest distance = $\sqrt{126 - (\frac{446}{\sqrt{1674}})^2} \approx 2.678$
I have done it a few times now and keep getting this answer; however, my provided answer is $\sqrt{6} \approx 2.449$. Where am I going wrong?