## Problem Statement

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

$\frac{x-1}{6}=\frac{y-3}{1}=\frac{z+5}{3}$

$3 x-2 y+5 z-3=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{6}=\frac{y-3}{1}=\frac{z+5}{3}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+6 t \\
y=3+t \\
z=-5+3 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$3(1+6 t)-2(3+t)+5(-5+3 t)-3=0$

$3+18 t-6-2 t-25+15 t-3=0$

$-31+31 t=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+6 \cdot 1=7 \\
y=3+1=4 \\
z=-5+3 \cdot 1=-2
\end{array}\right.
$$

We get:

$(7 ; 4 ;-2)$

## Problem Kuznetsov Analytic Geometry $14-26$