How do I determine the intersection point of a line $l$ and a plane $V$

Question

The line $l$ passes trough $(1,-4,0)$ and $(-4,4,-2)$. The given plane is: $3x+4y-4z=37$.

The first thing I thought to do is writing the two points as a vector. I came up with $(-5,8,-2)$. I tried using the vector for the following parameter equation: $(-5,8,-2)+t(3,4,-4)$ but apparently this is wrong. Could someone help me out?

Answer 1

Direction vector: $\vec d=(-5,8,-2)$.

Line: $\vec r = (1,-4,0)+t (-5,8,-2)$.

Plane: $3x+4y-4z=37$;

Solve for $t$:

$3(1-5t)+4(-4+8t)-4(-2t)=37;$

And then?