## Problem Statement

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

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

$3 x+7 y-5 z-11=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$3(5-t)+7(-3+5 t)-5(1+2 t)-11=0$

$15-3 t-21+35 t-5-10 t-11=0$

$-22+22 t=0$

$t=1$

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

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

We get:

$(4 ; 2 ; 3)$

Problem Kuznetsov Analytical Geometry 14-22