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Let the foot of the perpendicular from the point $\mathrm{1,2,4}$ on the line $\mathrm{\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}}$ be $\mathrm{P}$ . Then the distance of $\mathrm{P}$ from the plane $\mathrm{3 x+4 y+12 z+23=0}$ is
Option: 1
$5$

Option: 2
$\frac{50}{13}$

Option: 3
$4$

Option: 4
$\frac{63}{13}$

Answers (1)

best_answer

Let the foot of perpendicular be

$\mathrm{P(4 \lambda-2,2 \lambda+1,3 \lambda-1)} \\$

$\mathrm{A:(1,2,4) \quad A P \perp L} \\$

$\mathrm{\Rightarrow(4 \lambda-2-1,2 \lambda+1-2,3 \lambda-1-4)-(4,2,3)=0 }\\$

$\mathrm{\Rightarrow 4(4 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-5)=0} \\$

$\mathrm{\Rightarrow 29 \lambda=29 \Rightarrow \lambda=1} \\$

$\mathrm{\Rightarrow P(2,3,2) }$

Distance of $\mathrm{P }$ from $\mathrm{3 x+4 y+12 z+23=0 }$ is

$\mathrm{ d =\frac{3 \times 2+4 \times 3+2 \times 12+23}{\sqrt{3^{2}+4^{2}+12^{2}}} }\\$

$\mathrm{=\frac{65}{13}=5}$

Hence the correct answer is option 1

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