## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{8}=\frac{y-8}{-5}=\frac{z+5}{12}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+8 t \\
y=8-5 t \\
z=-5+12 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$$
\begin{aligned}
& (1+8 t)-2(8-5 t)-3(-5+12 t)+18=0 \\
& 1+8 t-16+10 t+15-36 t+18=0 \\
& -18 t+18=0 \\
& t=1
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
x=1+8 \cdot 1=9 \\
y=8-5 \cdot 1=3 \\
z=-5+12 \cdot 1=7
\end{array}\right.
$$

We get:

$(9 ; 3 ; 7)$

## Problem Kuznetsov Analytic Geometry 14-20
## Problem Statement

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

$\frac{x-1}{8}=\frac{y-8}{-5}=\frac{z+5}{12}$

$x-2 y-3 z+18=0$