## Problem Statement

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

$\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z+5}{0}$

$x+7 y+3 z+11=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-3}{1}=\frac{y-1}{-1}=\frac{z+5}{0}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=3+t \\
y=1-t \\
z=-5
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(3+t)+7(1-t)+3 \cdot(-5)+11=0$ $3+t+7-7 t-15+11=0$ $-6 t+6=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=3+1=4 \\
y=1-1=0 \\
z=-5
\end{array}\right.
$$

We get:

$(4 ; 0 ;-5)$

## Problem Kuznetsov Analytical Geometry 14-21