## Problem Statement

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

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

$x+3 y+5 z-42=0$
## Solution

Let's write the parametric equations of the line.

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

Substitute into the equation of the plane:

$(2+2 t)+3(2-t)+5(4+3 t)-42=0$

$2+2 t+6-3 t+20+15 t-42=0$

$14 t-14=0$

$t=1$

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

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

We get:

$(4 ; 1 ; 7)$

## Problem Kuznetsov Analytic Geometry 14-15