## Problem Statement

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

$\frac{x-2}{4}=\frac{y-1}{-3}=\frac{z+3}{-2}$

$3 x-y+4 z=0$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-2}{4}=\frac{y-1}{-3}=\frac{z+3}{-2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=2+4 t \\
y=1-3 t \\
z=-3-2 t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

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

$6+12 t-1+3 t-12-8 t=0$

$7 t-7=0$

$t=1$

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

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

We get:

$(6 ;-2 ;-5)$

## Problem Kuznetsov Analytic Geometry 14-27