c)

r = (2 - t)i - j + tk, t ∈ R

So
r = (2-t -1 t)
r = (2 -1 0) + t(-1 0 1)

This line is parallel to (-1 0 1)

So the line you want to find is also parallel to (-1 0 1)


(3 1 4) is a point on the line.


So the equation of the line can be :

r = (3 1 4) + λ(-1 0 1), λ ∈ R

Good observation for part d) that both lines pass through the same point.

i.e they intersect there.

Since the line of interest is perpendicular to both of these lines, it means that the line is also parallel to the cross product of those two lines.

So just find their cross product.