## Problem Statement

Find the point of intersection of the line and the plane.
$\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-8}{0}$

$5 x+9 y+4 z-25=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-3}{1}=\frac{y+2}{-1}=\frac{z-8}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+t \\
y=-2-t \\
z=8
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$5(3+t)+9(-2-t)+4 \cdot 8-25=0$

$15+5 t-18-9 t+32-25=0$

$-4 t+4=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=3+1=4 \\
y=-2-1=-3 \\
z=8
\end{array}\right.
$$

We get:

$(4 ;-3 ; 8)$

## Problem Kuznetsov Analytic Geometry 14-24