## Problem Statement

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

$\frac{x+3}{2}=\frac{y-1}{3}=\frac{z-1}{5}$

$2 x+3 y+7 z-52=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x+3}{2}=\frac{y-1}{3}=\frac{z-1}{5}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=-3+2 t \\
y=1+3 t \\
z=1+5 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$$
\begin{aligned}
& 2(-3+2 t)+3(1+3 t)+7(1+5 t)-52=0 \\
& -6+4 t+3+9 t+7+35 t-52=0 \\
& 48 t-48=0 \\
& t=1
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
x=-3+2 \cdot 1=-1 \\
y=1+3 \cdot 1=4 \\
z=1+5 \cdot 1=6
\end{array}\right.
$$

We get:

$(-1 ; 4 ; 6)$

## Problem Kuznetsov Analytic Geometry $14-17$