Oh sorry Im dumb, I realized another thing, since I have two points on the line I should be able to find the direction vector of the incoming line? I forgot that (-4, -5, 1) is one point, (5, 1, 4) is the point of intersection with the plane. Assuming that (-4, -5, 1) is the starting point for the incoming line, as I believe we can assume since t = 0. (5, 1, 4) - (-4, -5, 1) = (9, 6, 3). Now we have the normal vector of the plane n (1, 2, -1) and the direction vector i (9, 6, 3).

r = i - 2((i · n)/n · n)n

i*n = (9,6,3)*(1,2,-1)=18

n*n = (1,2,-1)*(1,2,-1)=6

Input into formula r = (9,6,3) - 2(18/6)(1,2,-1) = (9,6,3)-(6,12,-6)=(3,-6,9)

using r = a+dt, I input a as the point of intersection with plane (5, 1, 4), and d as the direction vector I got (3, -6, 9). (x,y,z)=(5,1,4)+t(3,-6,9) which gives x=5+3t, y=1-6t, z=4+9t. so the parameterform becomes (x,y,z)=(5+3t, 1-6t, 4+9t)

Is there any way I can check if my calculations are correct? Or could you double check if what I've done seems reasonable? :)