Distance between point and plane Find the distance from the Point $A = (1,0,2)$ to the plane passing through the point $(1,-2,1)$ and perpendicular to the line given by the parametric equations: 
$$
\begin{align}
x & = 7, \\ 
y & = 1 + 2t, \\
z & = -3 + t.
\end{align}
$$
The answer is $\sqrt{5}$, but I can't seem to get that. I get that the plane equation ends up being $0x + 2y + z + 3 = 0$, but then when I try to compute the distance it turns out to be $\sqrt{3}/\sqrt{5}$ or something along those lines. 
 A: We will utilize the point-plane distance formula [1].
$$D=\frac{\left\lvert ax_0+by_0+cz_0+d \right\rvert}{\sqrt{a^2+b^2+c^2}}$$
We already know that the point in question is $A(1,0,2)$ which means $x_0=1$, $y_0=0$, and $z_0=2$.
$$D=\frac{\left\lvert a(1)+b(0)+c(2)+d \right\rvert}{\sqrt{a^2+b^2+c^2}}=\frac{\left\lvert a+2c+d \right\rvert}{\sqrt{a^2+b^2+c^2}}$$
Since the plane in question is orthogonal to the given line, we can find a normal vector to the plane that is contained in the line. The line contains the points $P(7,1,-3)$ and $Q(7,3,-2)$ when $t=0$ and $t=1$, respectively. The vector from $P$ to $Q$ can be found easily. It is orthogonal to the plane.
$$\vec{\mathbf{n}}=\langle(7-7),(3-1),(-2+3)\rangle=\langle0,2,1\rangle$$
We can place the components of this vector into the general equation of a plane in $\mathbb{R}^3$ [2].
$$0x+2y+1z+d=0\implies2y+z+d=0$$
This plane contains the point $(1,-2,1)$. We can solve for the constant $d$.
$$2(-2)+(1)+d=0\implies-3+d=0\implies d=3$$
We can now return to the distance formula and finish solving the problem with $a=0$, $b=2$, $c=1$, and $d=3$.
$$D=\frac{\left\lvert a+2c+d \right\rvert}{\sqrt{a^2+b^2+c^2}}=\frac{\left\lvert (0)+2(1)+(3)\right\rvert}{\sqrt{(0)^2+(2)^2+(1)^2}}=\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{5}=\sqrt{5}$$
