How to find the equation of a plane containing a given point and perpendicular to a given line? I am asked to find the plane that contains the point $(-4, 7, -3)$ and is perpendicular to the line defined by the parametric equations $x = -4 + t$, $y = 7 + 3t$ and $z = -2t$. Now, I know that if $n = (a, b, c)$ is a normal vector to a plane $P$ and if $r_{0} = (x_{0}, y_{0}, z_{0})$ is a point in the plane, then $$P = \{ r: n \cdot (r - r_{0}) = 0 \} = \{ (x, y, z): a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0 \},$$ so I thought I just had to take a particular point in the line and call it $n$, then call $r_{0} = (-4, 7, -3)$ and apply the aforementioned equation, but I think there is something wrong with this approach since the four possible answers to the exercise are:
\begin{equation}
x - 3y - 2z = 21
\end{equation}
\begin{equation}
3x +y - 2z = 23
\end{equation}
\begin{equation}
x + 3y - 2z = 23
\end{equation}
\begin{equation}
x + 3y +z = 21,
\end{equation}
leaving the possible choices of $n$ to $(1, -3, -2)$, $(3, 1, -2)$, $(1, 3, -2)$ or $(1, 3, 2)$. The problem is that none of those points are in the given line. 
I would like to know what is wrong with my strategy to find the equation of the plane.
 A: If the plane is perpendicular to the line, then the direction vector of the line is normal to the plane. The direction vector of your line is $\vec{n} = (1,3,-2)$. So, using $\vec{r}_0 = (-4,7,-3)$, the plane equation $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ becomes:
$$
(1,3,-2) \cdot (x+4, y-7, z+3) = 0
$$
This simplifies to
$$
x + 3y -2z = 23
$$
Your approach didn't work because you used a point on the line to define $\vec{n}$; you need to use the direction vector of the line, instead.
A: A more simplified method would be to sort what you are given first, so you are given a point, let us call this point P(-4,7,-3), and parametric equations for a line perpendicular to the line you are solving for
x= -4 + t
y= 7 + 3t
z= -2t
A simple trick for solving for the vector (n) is to simply look at the coefficients of t in each equation, these are the (a, b, c) of the vector.
Therefore let us say vector n = (1,3,-2). That is step 1.
Next, the line needed is perpendicular to the given line, so the assumption, to be perpendicular the vector given from the product of point P and another point, let us call it Q, needs to equal 0. The equation for vector PQ is also a(x-x0) + b(y-y0) + c(z-z0) which should equal 0.
Then by substituting in the solved vector n = (1,3,-2) and the given point P for (x0,y0,z0) = (-4,7-3) you get:
[1(x+4) + 3(y-7) - 2(z+3)] = 0
which simplified is:
[x + 4 + 3y - 21 - 2z - 6] = 0
[x + 3y - 2z - 23] = 0
Final Answer = x + 3y - 2z = 23
