Equation of plane parallel to a vector and containing two given points 
I'm not sure how to solve this. 
I started by finding the equation of the line AB. 

Hints:
Let the wanted plane be $\;m: ax+by+cz+d=0\;$ , then it must fulfill:
$$\begin{align}&0=(a,b,c)\cdot(1,-5,-3)\implies&a-5b-3c=0\\{}\\
&A\in m\implies&3a-2b+4c+d=0\\{}\\
&B\in m\implies&2a-b+7c+d=0\end{align}$$
Well, now solve the above linear system. Pay attention to the fact that you shall get a parametrized answer and you can choose values so as to get nice values for $\;a,b,c,d\;$ (i.e., you do not need a fourth independent equation).
The line through $\;A,\,B\;$ is $\;\ell:\;A+t\vec{BA}\;,\;\;t\in\Bbb R$ , so find the value of $\;t\;$ for which an element in this line minimizes the distance to $\;C\;$ (further hint: it must be an element  $\;H\in\ell\;$ perpendicular $\;\vec{HC}\;$) , or use one of the formulas to evaluate distance from point to line in the space.
