# Need solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 9 textbook solution.

Answer :  The answer of the given question is $\lambda=4$

Hint :

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

Given : Points: $(1,1,1) \text { and } x-y+z+\lambda=0 \text { be } 5$

Solution :

The distance of the point $(1,1,1)$ from the origin

We know, distance of $\left(x_{1}, y_{1}, z_{1}\right)$ from the origin is $\sqrt{x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}$

Putting values of $x_{1}, y_{1}, z_{1}=1$

Required distance $=\sqrt{3}$

Distance of the point  $(1,1,1)$ from plane $x-y+z+\lambda=0$

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

Is given by

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

Putting necessary values,

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

According to the question

\begin{aligned} &\left|\frac{1+\lambda}{\sqrt{3}}\right| \times \sqrt{3}=5 \\ &|1+\lambda|=5 \\ &1+\lambda=5 \text { Or } 1+\lambda=-5 \\ &\lambda=4 \text { Or } \lambda=-6 \end{aligned}

$\lambda = 4$ [$\lambda \neq -6$ Because distance cannot be negative]

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