# Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 4 textbook solution.

Answer : The answer of the given question is $x+2 y-2 z+4=0, x+2 y-2 z-8=0$

Hint :

$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

Given : Plane: $x+2 y-2 z+8=0$, which are at a distance of 2 units from the point $(2,1,1)$

Solution :

The planes are parallel to $x+2 y-2 z+8=0$ they must be of the form

$x+2 y-2 z+\theta=0$

We know, the distance of point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane $a x+b y+c z+d=0$ given by

$P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

According to the question, the distance of the plane from $(2,1,1)$ is 2 units

\begin{aligned} &\left|\frac{1 \times 2+2 \times 1+(-2) \times(1)+\theta}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}\right|=2 \\ &\left|\frac{2+\theta}{3}\right|=2 \end{aligned}

\begin{aligned} &\frac{2+\theta}{3}=2 \text { Or } \frac{2+\theta}{3}=-2 \\ &\theta=4 \text { Or } \theta=-8 \end{aligned}

Hence the required planes are

\begin{aligned} &x+2 y-2 z+4=0 \\ &x+2 y-2 z-8=0 \end{aligned}

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