Q:

Find an equation for a plane containing a line, orthogonal to another line

I was wondering if someone could give me directions on how to solve the following problem:

I've gotten the equations of the lines pretty easily, with the line that must be contained being represented as the following:
$$R_1=(0,1,-1)+T(1,3,2)$$
And the line to be orthogonal against being:
$$R_2=(1,-5,3)+U(-17,1,7) $$
But I'm stuck as to go from here. How do we define a plane using a line and a given point? How do we make that plane orthogonal to another line?

A:

The plane's normal vector is parallel to the direction vector of $R_2$, which means its equation is
$$(-17,1,7)\cdot(x,y,z)=k$$
To find $k$, note that the starting point of $R_1$ $(0,1,-1)$ is in the plane:
$$(-17,1,7)\cdot(0,1,-1)=-6$$
So the equation of the plane is
$$-17x+y+7z=-6$$
As verification, we can check that $R_1$ and $R_2$'s direction vectors are perpendicular ($(1,3,2)\cdot(-17,1,7)=0$), implying that the plane contains $R_1$.

