Does the line $(2,1,1)+t(-3,1,5)$ live within the plane $31x+3y+18z=62$? I have a doubt with this exercise:

Have the plane $$31x+3y+18z=62$$
What is the distance between this plane and some line $(x,y,z) = (2,1,1) + t(-3,1,5)$ for some $t\in\mathbb{R}$?

The answer is $0$. The book says that this line lives within the plane.

But here is the problem: if the line lives within the plane, then the point $(2,1,1)$ should belong to the plane as well. However, it doesn't seem to be the case:
$$31(2) + 3(1) + 18(1) = 62 + 3 + 18 = 83 \not = 62$$
So what's up?

I have managed to confirm that the plane is parallel with the line. You just get the normal vector $(31,3,18)$ and then calculate its dot product with $(-3,1,5)$. This yields $0$, so the plane is parallel to the line indeed. Given this fact, there are two scenarios: either the line lives within the plane, or it doesn't. If it did, then the point $(2,1,1)$ should be in the plane! I think...
 A: The line $X(t) = (2-3t,1+t,1+5t)$ must satisfy the plane equation for all values of $t$. We have: $$31(2-3t) + 3(1+t) + 18(1+5t) = 83 \neq 62,$$
so the line isn't in the plane.
The normal vector to the plane is $(31,3,18)$. So, take a point in the line, say, $(2,1,1)$. Consider the line $Y(\lambda) = (2,1,1) + \lambda(31,3,18)$. Let's find the point where this line crosses the plane. We have $Y(\lambda) = (2+31\lambda,1+3\lambda,1+18\lambda)$. We must have: $$31(2+31\lambda) + 3(1+3\lambda)+ 18(1+18\lambda) = 62 \\ 83 + 1294\lambda = 62 \implies \lambda = -\frac{21}{1294}$$
So the distance would be the distance from $(2,1,1)$ to $Y(-21/1294)$. Which is very strange.
A: Let's find the intersection points of the plane and the line.
Let $M(x,y,z)$ a point in the plane and in the line.
So $31x+3y+18z=62$ and $\exists t\in\mathbb{R},\,\left \{
\begin{array}{c @{=} c}
    x=2-3t \\
    y=1+t  \\
    z=1+5t
\end{array}
\right.$
So $31(2-3t)+3(1+t)+18(1+5t)=62$ which means that $83=62$ which is a contradiction.
So the intersection of the plane and the line is $\emptyset$ which means that this line is parallel to the plane.
Let's take a point from the line, for example the point $(2,1,1)$.
The formula of the distance between a point $A(x_A,y_A,z_A)$ and a plane $(P):\, ax+by+cz+d=0$ is: $$d_{A,P}=\dfrac{|ax_A+by_A+cz_A+d|}{\sqrt{a^2+b^2+c^2}}$$
So if you use it you'll get the distance between the point $(2,1,1)$ and the given plane. In fact the line is parallel to the plane so the distance of two points of the line to the plane is the same and so the distance between the given line and the given plane is the distance between $(2,1,1)$ and the given plane.
