## Problem Statement

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

$\frac{x+1}{-2}=\frac{y}{0}=\frac{z+1}{3}$

$x+4 y+13 z-23=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+1}{-2}=\frac{y}{0}=\frac{z+1}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1-2 t \\
y=0 \\
z=-1+3 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(-1-2 t)+4 \cdot 0+13(-1+3 t)-23=0$

$-1-2 t-13+39 t-23=0$

$37 t-37=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=-1-2 \cdot 1=-3 \\
y=0 \\
z=-1+3 \cdot 1=2
\end{array}\right.
$$

We get:

$(-3 ; 0 ; 2)$

Kuznetsov Analytical Geometry 14-25