## problem statement

Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point

$A$ belongs to the image of plane $a$?

$A(1 ; 1 ; 1)$

$a: 7 x-6 y+z-5=0$

$k=-2$
## Solution

When transforming similarity with the center at the origin of the plane

$a: A x+B y+C z+D=0$ and the coefficient $k$, the plane transitions to

$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:

$a^{\prime}: 7 x-6 y+z+10=0$

Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:

$7 \cdot 1-6 \cdot 1+1+10=0$

$7-6+1+10=0$

$12=0$

Since $12 \neq 0$, point $A$ does not belong to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-17