## Problem Statement

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

$\frac{x+3}{0}=\frac{y-2}{-3}=\frac{z+5}{11}$

$5 x+7 y+9 z-32=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+3}{0}=\frac{y-2}{-3}=\frac{z+5}{11}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3 \\
y=2-3 t \\
z=-5+11 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$5 \cdot(-3)+7(2-3 t)+9(-5+11 t)-32=0$

$-15+14-21 t-45+99 t-32=0$

$78 t-78=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=-3 \\
y=2-3 \cdot 1=-1 \\
z=-5+11 \cdot 1=6
\end{array}\right.
$$

We get:

$$
(-3 ;-1 ; 6)
$$

## Problem Kuznetsov Analytic Geometry 14-30