Example 1.24. Perpendiculars are dropped from the point $P(2 ; 3 ;-5)$ to the coordinate planes. Find the equation of the plane passing through their feet.
Solution. The bases of the perpendiculars dropped onto the coordinate planes will be the following points $M_{1}(2 ; 3 ; 0), M_{2}(2 ; 0 ;-5)$, $M_{3}(0 ; 3 ;-5)$. To write the equation of the plane passing through points $M_{1}, M_{2}, M_{3}$, we will use the equation

$$
\left|\begin{array}{rrr}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|=0
$$

we find

$$
\left|\begin{array}{rrr}
x-2 & y-3 & z \\
0 & -3 & -5 \\
-2 & 0 & -5
\end{array}\right|=0
$$

or

$$
15 x+10 y-6 z-60=0
$$