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Let the line passing through the point P(2,-1,2) and Q(5,3,4) meet the plane x-y+z=4 at the point T.Then the distance of the point R from the plane x+2 y+3 z+2=0 measured parallel to the line \frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}is equal to

Option: 1

3


Option: 2

\sqrt{61}


Option: 3

\sqrt{31}


Option: 4

\sqrt{189}


Answers (1)

best_answer

Line: \frac{x-5}{3}=\frac{y-3}{4}=\frac{z-4}{2}=\lambda
\mathrm{R}(3 \lambda+5,4 \lambda+3,2 \lambda+4)

\therefore 3 \lambda+5-4 \lambda-3+2 \lambda+4=4

\lambda+6=4 \quad \therefore \lambda=-2

\therefore \mathrm{R} \equiv(-1,-5,0)

Line : \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+5}{2}=\frac{\mathrm{z}-0}{1}=\mu
Point \: \: \mathrm{T}=(2 \mu-1,2 \mu-5, \mu)

It lies on plane
2 \mu-1+2(2 \mu-5)+3 \mu+2=0
\mu=1
\therefore \mathrm{T}=(1,-3,1)
\therefore \mathrm{RT}=3

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