## Problem Statement

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

$3 x-7 y-2 z+7=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$3(1+t)-7 \cdot 3-2(-2-2 t)+7=0$

$3+3 t-21+4+4 t+7=0$

$7 t-7=0$

$t=1$

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

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

We get:

$(2 ; 3 ;-4)$

## Problem Kuznetsov Analytic Geometry 14-29