Find a line that is perpendicular to a plane that passes through a point This is a homework question I am having difficulty answering. (It is Calc III, but this problem only concerns vectors).

(A) Find the parametric equations for the line through the point $P =
 (4, -1, 4)$ that is perpendicular to the plane $3x + 1y  - 5z = 1$. Use
  "$t$" as your variable, $t = 0$ should correspond to $P$, and the velocity
  vector of the line should be the same as the normal vector to the
  plane found directly from its equation.
(B) At what point $Q$ does this line intersect the $yz$-plane?

So Here's what I've tried conceptualizing. First we have point $P$, which we can think of as a vector from the origin. We can also create the normal vector from the plane: $n = \langle 3, 1, -5\rangle$. We can pick a point on the plane that satisfies the equation $p_0 = (2,0,1)$. We can now use this information to start forming a triangle we can solve for. One side: $n - p_0 = \langle 1, 1, -6\rangle$. The hypotenuse can be found with $P - p_0 = \langle 2, -1, 3\rangle$. Which leaves us with one remaining side, which will be a vector within the plane which should give us the point in which the line will intersect with the plane, which we can turn into a vector from the origin and add to the unit vector of $n$ to create our equation.
I think finding this remaining side should be how we get all this information, but I don't know if that's right or if that's even the correct approach. Can anyone give me any advice?
 A: If $\mathbf{n}$ is the normal vector to the given plane and $\mathbf{p}$ is the point through which the line is supposed to pass, then the equation of the line will be of the form $\mathbf{r}=\mathbf{p}+t\mathbf{n}$. You already have both those vectors so you don't need anything else.
A: A
We can easily read off a normal vector to the plane, as the $x$ component will be the coefficient of $x$ in the Cartesian form, the $y$ component will be the coefficient of $y$ in the Cartesian form, and so on. Therefore the normal vector to the plane is given by $$ \vec{n}=<3,1,-5>.$$
We know that any line which is perpendicular to the plane must run parallel to $\vec{n}$. Given that we want to find the parametrization (with parameter $t$) of such a line that passes through $P$ at $t=0$, the line is given by $$\vec{r}(t)=<4,-1,4> + t<3,1,-5>.$$
B
The $yz$-plane in Cartesian form is $x=0$. So, we want to find for what value of $t$ does the first component of $\vec{r}(t)$ equal to $0$.
$$4 + 3t=0 \implies t=-{4 \over 3}$$
Then substituting this value of $t$ into our parametrization of the line yields
$$\vec{r}\left(-{4 \over 3}\right)= <4,-1,4> -{4 \over 3}<3,1,-5>=<0,-{7\over 3},{32 \over 3}>.$$
So, $Q=\left(0,-{7\over 3},{32 \over 3}\right)$.
A: For (B) you solve
$$
0 = 4 + 3 t
$$
for $t$. Thus $t = -4/3$. And then insert this into the line equation to get the point.
