## Problem Statement

Find the point of intersection of the line and the plane.

$\frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+4}{-1}$

$2 x-y+3 z+23=0$
## Solution

Let's write the parametric equations of the line.

$\frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+4}{-1}=t \Rightarrow$

$\left\{\begin{array}{l}x=-2-t \\ y=1+t \\ z=-4-t\end{array}\right.$

Substitute into the equation of the plane:

$2(-2-t)-(1+t)+3(-4-t)+23=0$

$-4-2 t-1-t-12-3 t+23=0$

$-6 t+6=0$

$t=1$

Find the coordinates of the intersection point of the line and the plane:

$$
\left\{\begin{array}{l}
x=-2-1=-3 \\
y=1+1=2 \\
z=-4-1=-5
\end{array}\right.
$$

We get:

$(-3 ; 2 ;-5)$

Problem Kuznetsov Analytic Geometry 14-9