Prove some equations about distances between points and planes

Prove that the distance from the point $(x_0, y_0, z_0)$ to the plane
$ax + by + ca + d = 0$

is given by the formula

$\frac{| ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}.$
Find the point $P$ on the plane $5x - 14y + 2z + 9 = 0$ which is nearest to the point $Q = (-2,15,-7)$ .

Proof. By Theorem 13.6 (page 476 of Apostol) we know that the distance from a point to a plane is given by
$\begin{align*} d &= \frac{| N \cdot P |}{\lVert N \rVert} \\[9pt] &= \frac{ | (a,b,c) \cdot (x_0, y_0, z_0) + d|}{\sqrt{a^2+b^2+c^2}} \\[9pt] &= \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2+b^2+c^2}}. \qquad \blacksquare \end{align*}$
A normal to the plane is given by $N = (5,-14,2)$ . So, $P \cdot N = 9$ for any point $P \in M$ . Further, the distance from $M$ to a point $Q$ not on $M$ is minimal when $X = tN$ where
$t = \frac{P \cdot N}{N \cdot N} = \frac{9}{225} = \frac{1}{25}.$

Thus,

$X = \frac{1}{25} (5,-14,2).$

Naming $X$ to be the point $P$ we have

$P = \frac{1}{25} (5,-14,2).$

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