The equations of planes parallel to the plane $x+2 y+2 z+8=0$, which are at a distance of 2 units from the…
The equations of planes parallel to the plane $x+2 y+2 z+8=0$, which are at a
distance of 2 units from the point $(1,1,2)$ are
$x+2 y+2 z-13=0$ or $x+2 y+2 z-1=0$
$x+2 y+2 z-6=0 \quad$ or $\quad x+2 y+2 z-7=0$
$x+2 y+2 z+3=0 \quad$ or $\quad x+2 y+2 z-5=0$
$x+2 y+2 z-5=0 \quad$ or $\quad x+2 y+2 z-3=0$
Solution
The equation of the plane parallel to the plane $x+2 y+2 z+18=0$ is $x+2 y+2 z+\lambda=0$. Now, the distance of this plane from the print $(1,1,2)$ is
$\therefore\left|\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^{2}+2^{2}+2^{2}}}\right|=\left|\frac{7+\lambda}{3}\right|$
$\begin{array}{l}
\text { We have }\left|\frac{7+\lambda}{3}\right|=2 \Rightarrow \frac{7+\lambda}{3}=\pm 2 \\
\therefore \lambda=\pm 6-7=1 \Rightarrow \lambda=-1 \text { or } \lambda=-13
\end{array}$
Hence, equations of plane are $x+2 y+2 z-1=0$ or $x+2 y+2 z-13=0$