Question
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
A.
$$0$$
B.
$$1$$
C.
$$\sqrt 2 $$
D.
$$2\sqrt 2 $$
Answer :
$$2\sqrt 2 $$
Solution :
The equation of plane through the point $$\left( {1,\, - 2,\,1} \right)$$ and perpendicular to the planes $$2x-2y+z=0$$ and $$x-y+2z=4$$ is given by
\[\left| \begin{array}{l}
x - 1\,\,\,\,\,y + 2\,\,\,\,\,z - 1\\
\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,2
\end{array} \right| = 0\,\, \Rightarrow x + y + 1 = 0\]
It’s distance from the point (1, 2, 2) is
$$\left| {\frac{{1 + 2 + 1}}{{\sqrt 2 }}} \right| = 2\sqrt 2 .$$