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

Answer : The answer of the given question is $\frac{12}{\sqrt{29}}$ units

Hint :

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

Given : Plane makes intercept -6,3,4 respectively on the coordinate axes

Solution :

The equation of the plane which makes intercept a, b and c with the x, y and z axes respectively is

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Putting the values of a, b and c

Require equation of the plane

\begin{aligned} &\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1 \\ &-2 x+4 y+3 z=12 \end{aligned}

We know, the distance of the point $\left(x_{1}, y_{1}, z_{1}\right)$ from the plane

$\pi: a x+b y+c z+d=0$ Is given by

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

Distance from the origin i.e. $(0,0,0):\left|\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|$

Required distance

\begin{aligned} &=\left|\frac{-12}{\sqrt{(-2)^{2}+4^{2}+3^{2}}}\right| \\ &=\frac{12}{\sqrt{29}} \end{aligned}

The length of the perpendicular from the origin on the plane $=\frac{12}{\sqrt{29}} \text { Units }$

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