## Problem Statement

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

$\frac{x+1}{3}=\frac{y-3}{-4}=\frac{z+1}{5}$

$x+2 y-5 z+20=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+1}{3}=\frac{y-3}{-4}=\frac{z+1}{5}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-1+3 t \\
y=3-4 t \\
z=-1+5 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$(-1+3 t)+2(3-4 t)-5(-1+5 t)+20=0$

$-1+3 t+6-8 t+5-25 t+20=0$

$-30 t+30=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=-1+3 \cdot 1=2 \\
y=3-4 \cdot 1=-1 \\
z=-1+5 \cdot 1=4
\end{array}\right.
$$

We get:

$(2 ;-1 ; 4)$

## Problem Kuznetsov Analytic Geometry $14-2$