## Problem Statement

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

$$
\begin{aligned}
& \frac{x-5}{1}=\frac{y-3}{-1}=\frac{z-2}{0} \\
& 3 x+y-5 z-12=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

$\frac{x-5}{1}=\frac{y-3}{-1}=\frac{z-2}{0}=t \Rightarrow$

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

Substitute into the equation of the plane:

$3(5+t)+(3-t)-5 \cdot 2-12=0$

$15+3 t+3-t-10-12=0$

$2 t-4=0$

$t=2$

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

$$
\left\{\begin{array}{l}
x=5+2=7 \\
y=3-2=1 \\
z=2
\end{array}\right.
$$

We get:

$(7 ; 1 ; 2)$

## Problem Kuznetsov Analytic Geometry 14-5