Does the point $(1,8,3)$ line on the line with parametric equation: $$x = 5 + 2t$$ $$y = 2 + 6t$$ $$z = 1 + 3t$$
I know how to solve if they give me a equation of a plane and ask whether the line intersect the plane or not. If it intersect I can find where it intersect(the point of intersection). Can someone help me please!
On
If you want to know whether a given point $(x,y,z)$ belongs to this line, you are asking whether \begin{align*} x-5&=2t\\ y-2&=6t\\ z-1&=3t \end{align*} i.e., whether $(x-5,y-2,z-1)=(2,6,3)t$ for some $t\in\mathbb R$.
In the other words, you want to know whether the vector $(x-5,y-2,z-1)$ is a multiple of the vector $(2,6,3)$.
For $(x,y,z)=(1,8,3)$ you get $(x-5,y-2,z-1)=(-4,6,2)$. Is it a multiple of $(2,6,3)$?
HINT:
Putting the values of the $x,y$ and $z$ co-ordinates of the point $(1,8,3)$ in the parametric equation of the line, see if they yield the same value of $t$. If yes, then the point is on the line. Else not.