Distance from a point to a plane (in the parametric form not general form) I have a point $Q = [7 2 1]$ and I need to find the distance to the plane $P = [1 2 0] + s[1 -1 1] + t[1 0 3]$. I'm not sure how to approach this question.I know to do this question if the equation of the plane was given in the form $Ax + By + Cz = D$ but I'm not sure how to do it in this form
I have thought of two methods but I'm not sure which is correct.
This is the first methodand this is the second method
Could someone please advise me if either method is correct and if not the correct method.
Thanks

A point in the plane is $(1,2,0)$ and a vector normal to the plane is 
\begin{align*}(1,-1,1) \times (1,0,3) & = (-3,-2,1).
\end{align*}
So the equation of the plane is $$-3(x-1) -2(y-2) +z = 0$$ which simplifies to $$3x +2y - z = 7.$$ You can take it from here. Edit: both of your methods are equivalent, but you made a mistake in the second attempt. 
