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  • Let be the plane, passing through the point <img alt="(1,-1,-5)" src="https://entrancecorner.oncodeco

Let \mathrm{P} be the plane, passing through the point (1,-1,-5) and perpendicular to the line joining the points (4,1,-3)$ and $(2,4,3). Then the distance of \mathrm{P} from the point (3,-2,2) is
 

Option: 1

5


Option: 2

4


Option: 3

7


Option: 4

6


Answers (1)

best_answer

Let \mathrm{A}(4,1,-3) \& \mathrm{~B}(2,4,3)
\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{AB}}=(-2,3,6)
Plane \mathrm{ P} is :
\begin{aligned} & -2(\mathrm{x}-1)+3(\mathrm{y}+1)+6(\mathrm{z}+5)=0 \\ & -2 \mathrm{x}+2+3 \mathrm{y}+3+6 \mathrm{z}+30=0 \\ & 2 \mathrm{x}-3 \mathrm{y}-6 \mathrm{z}=35 \end{aligned}
Distance of \mathrm{P} from the point (3,-2,2) is
\begin{aligned} & =\frac{|6+6-12-35|}{\sqrt{2^2+3^2+6^2}} \\ & =\frac{35}{7}=5 \text { Ans. (1) } \end{aligned}

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