Let the foot of the perpendicular from the point (1,2,4) on the line (x+2)/(4)=(y-1)/(2)=(z+1)/(3) be P . Then the distance of P from the plane 3x+4y+12z+23=0 is

3D Geometry: Let the foot of the perpendicular from the point (1,2,4) on the line… | JEE Main 2022

JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1

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

3D Geometry

$5$✓ Correct answer$\frac{50}{13}$$4$$\frac{63}{13}$

Solution

$Q=(-2+4t,1+2t,-1+3t)$, $(Q-A)\cdot(4,2,3)=0$ with $A=(1,2,4)$: $4(4t-3)+2(2t-1)+3(3t-5)=0\Rightarrow29t-29=0$, $t=1$, foot $(2,3,2)$. Distance to $3x+4y+12z+23=0$: $\frac{6+12+24+23}{13}=\frac{65}{13}=5$. Option (A).

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