Question

A plane passes through the point $(1,-2,3)$ and is parallel to the plane $2x-2y+z=0$. The distance of the point $(-1,2,0)$ form the plane is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is D 5

Let equation of the plane passing through ($1,-2,3$) be

$A(x-1)+B(y+2)+C(z-3)=0$ $.......\left(1\right)$

Since this plane is parallel to the plane $2x-2y+z=0$

so, their normal will be parallel

therefore, $\frac{A}{2}=\frac{B}{-2}=\frac{C}{1}=K$

$A=2K,B=-2K,C=K$

Putting values of A, B and C in (1) we get

$2K(x-1)-2K(y+2)+K(z-3)=0$

$2(x-1)-2(y+2)+(z-3)=0$

$2x-2-2y-4+z-3$

$2x-2y+z-9=0$

Distance of point (-1,2,0) from this plane is

$d=\left|\frac{2(-1)-2\left(2\right)+0-9}{\sqrt{{\left(2\right)}^2+{(-2)}^2+{\left(1\right)}^2}}\right|$

$=\left|\frac{-2-4-9}{\sqrt{4+4+1}}\right|$

$=\frac{15}{3}$

$=5$

Hence, Option $D$ is the correct answer.