Condition of the problem

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

$$
\begin{aligned}
& \frac{x-1}{7}=\frac{y-2}{1}=\frac{z-6}{-1} \\
& 4 x+y-6 z-5=0
\end{aligned}
$$
## Solution

Let's write the parametric equations of the line.

$$
\begin{aligned}
& \frac{x-1}{7}=\frac{y-2}{1}=\frac{z-6}{-1}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=1+7 t \\
y=2+t \\
z=6-t
\end{array}\right.
\end{aligned}
$$

Substitute into the equation of the plane:

$4(1+7 t)+(2+t)-6(6-t)-5=0$

$4+28 t+2+t-36+6 t-5=0$

$35 t-35=0$

$t=1$

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

$$
\left\{\begin{array}{l}
x=1+7 \cdot 1=8 \\
y=2+1=3 \\
z=6-1=5
\end{array}\right.
$$

We get:

$(8 ; 3 ; 5)$

## Problem Kuznetsov Analytic Geometry 14-23