<s>[INST] How to find the plane which contains a point and a line I know that $\Pi$ contains the point $(2,0,5)$ and the line $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$. How would I find the minimum vector connecting the point and the line so I can then work out the cross product and thus the normal and equation of plane $\Pi$.
 [/INST] The point $A = (2,0,5)$ is in the plane,
If we set $x = 10$ into $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$ we get $B = (10,3,7)$ is in the plane.
If we set $x = 13$ into $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$ we get $C = (13,5,9)$ is in the plane.
$\quad \vec u =B - A = (8,3,2)$
$\quad \vec v =C - A = (11,5,4)$
Taking $\vec u \times \vec v$ we get that the equation has the form $-2x + 10y -7z = d$.
Plugging in the point $A$ we get the $d = -39$, so the answer is
$$\tag 1 -2x + 10y  -7z = -39$$
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