Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 13 maths textbook solution.

Answer : $2\sqrt{2}$

Hint :

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

Given : We have to find the distance of the point $(1,-2,4)$ from the plane passing through the point $(1,2,2)$

Solution : Let us consider the equation of plane passing through the point $(1,2,2)$ be

$a(x-1)+b(y-2)+c(z-2)=0$                      (1)

The direction ratios of the normal to the plane are a, b and c

The equation of the plane are

\begin{aligned} &x-y+2 z=3 \\ &2 x-2 y+z+12=0 \end{aligned}

Since the plane passing through the point $(1,2,2)$ is perpendicular the given plane

Therefore,

\begin{aligned} &a-b+2 c=0\\ &2 a-2 b+c=0 \end{aligned}

Now, eliminating a, b and c from (1), (2) and (3)

We get

\begin{aligned} &\left|\begin{array}{ccc} x-1 & y-2 & z-2 \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{array}\right|=0 \\ &(x-1)(-1+4)-(y-2)(1-4)+(z-2)(-2+2)=0 \end{aligned}

\begin{aligned} &3(x-1)+3(y-2)+0=0 \\ &3 x-3+3 y-6=0 \\ &3 x+3 y-9=0 \\ &x+y-3=0 \end{aligned}

Now using distance formula

The distance of the point $(1,-2,4)$ from the plane $x+y-3=0$ is

\begin{aligned} &=\left|\frac{1-2-3}{\sqrt{1^{2}+1^{2}+0^{2}}}\right| \\ &=\left|\frac{-4}{\sqrt{2}}\right| \\ &=2 \sqrt{2} \text { Units } \end{aligned}

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