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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 5 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 5 Maths Textbook Solution.

Answers (1)

Answer:Therefore ,the distance is 13 units

Hint: Use distance formula \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

Given:

\begin{aligned} &\text { Point } P(-1,-5,-10) \\ &A(2,-1,2) \& B(5,3,4) \\ &\text { and } x-y+z=5 \end{aligned}

Solution:Equation \: of \: line\: passing \: through (2,-1,2) \&(5,3,4) is

\frac{x-2}{5-2}=\frac{y+1}{3+1}=\frac{z-2}{4-2}

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=m

x=3 m+2, y=4 m-1, z=2 m+2

Now \: putting \: the \: value \: of \: \mathrm{x}, \mathrm{y} and \mathrm{z} in the \: equation \: of \: the \: palne \: x-y+z=5

(3 m+2)-(4 m-1)+(2 m+2)=5

m=0

So, the\: point \: of\: intersection \: of \: the \: line\: and\: the \: plane\: is (2,-1,2)

\therefore The\; distance \: of\: the \: point\; \mathrm{P}(-1,-5,-10)

=\sqrt{3^{2}+4^{2}+12^{2}}=13 \: unit

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