## Problem Statement

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

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

$x+3 y-5 z+9=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$(-1-3 t)+3(-2+2 t)-5(3-2 t)+9=0$

$-1-3 t-6+6 t-15+10 t+9=0$

$13 t-13=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=-1-3 \cdot 1=-4 \\
y=-2+2 \cdot 1=0 \\
z=3-2 \cdot 1=1
\end{array}\right.
$$

We get:

$(-4 ; 0 ; 1)$

## Problem Kuznetsov Analytic Geometry 14-6