Find a plane that is parallel to both $$\frac{x-1}2 = y = z+1$$ and $$x=\frac{y+1}3=\frac{z-1}4$$ and goes through the origin. Is it possible to place it so that it is equidistant to both lines?
I started with a cross product of the two vectors $\langle 2,1,1\rangle$ and $\langle1,3,4\rangle$ and then plugged in my point $(0,0,0)$ to get the plane $x-7y+5z=0$. Very unsure if this is the right method since I did a similar method trying to find a perpendicular plane to two planes that go through the origin.
Yes, your arithmetic is correct. As you discovered, the lines have direction vectors $(2,1,1)$ and $(1,3,4)$. The plane normal must be perpendicular to both of these, therefore parallel to their cross product, which is $(1,-7,5)$. The plane through the original with normal vector $(1,-7,5)$ can be described by the equation $x-7y+5z=0$.
You imply that this felt wrong because it's very similar to the computations you did to find a plane that's perpendicular to two give ones. But it's ok. The calculations should be similar: if a plane is perpendicular to two given planes, then it's parallel to both of their normal vectors. So, in both problems, the basic calculation is to find a plane that's parallel to two vectors. In the first problem, the two vectors are the direction vectors of the two lines; in the second problem, the two vectors are the normals of the two given planes. But, once you have the two vectors, the rest of the calculations are the same.