## Problem Statement

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

$\frac{x-1}{1}=\frac{y+1}{0}=\frac{z-1}{-1}$

$3 x-2 y-4 z-8=0$
## Solution

Let's write the parametric equations of the line.

$\frac{x-1}{1}=\frac{y+1}{0}=\frac{z-1}{-1}=t \Rightarrow$

$\left\{\begin{array}{l}x=1+t \\ y=-1 \\ z=1-t\end{array}\right.$

Substitute into the equation of the plane:

$3(1+t)-2(-1)-4(1-t)-8=0$

$3+3 t+2-4+4 t-8=0$

$7 t-7=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+1=2 \\
y=-1 \\
z=1-1=0
\end{array}\right.
$$

We get:

$(2 ;-1 ; 0)$

## Problem Kuznetsov Analytic Geometry 14-12