## Problem Statement

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

$$
\begin{aligned}
& \frac{x+2}{-1}=\frac{y-1}{1}=\frac{z+3}{2} \\
& x+2 y-z-2=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

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

$-2-t+2+2 t+3-2 t-2=0$

$-t+1=0$

$t=1$

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

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

We get:

$(-3 ; 2 ;-1)$

## Problem Kuznetsov Analytic Geometry 14-13