Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 5 textbook solution.

Answer:  The answer of the given question is that the points are equidistant from the plane

Hint :

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

Given :

Points $A(1,1,1) \text { and } B(-3,0,1)$

Plane $p=3 x+4 y-12 z+13=0$

Solution :

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

Distance of $(1,1,1)$ from the plane $=\left|\frac{3 \times 1+4 \times 1+(-12) \times 1+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|=\frac{8}{13} \text { Units }$

Distance of $(-3,0,1)$ from the plane $=\left|\frac{3 \times(-3)+4 \times 0+(-12) \times 1+13}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|=\frac{8}{13} \text { Units }$

The points  $(1,1,1) \&(-3,0,1)$ are equidistant from the plane  $3 x+4 y-12 z+13=0$

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