Question

Let $L_1:\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{3},$ $L_2:\frac{x-1}{-1}=\frac{y-2}{3}=\frac{3(z-3)}{5}$ and $L_3:\frac{x-1}{-32}=\frac{y-2}{-19}=\frac{z-3}{15}$ be three lines.
A plane is intersecting these lines at $A,B$ and $C$ respectively such that $PA=2,$ $PB=3$ and $PC=6$ where $P\equiv (1,2,3).$ If $V$ is the volume of the tetrahedron $PABC$ and $d$ is the perpendicular distance of the plane from the point $P,$ then

• A. $V=18$ cubic units
• B. $V=6$ cubic units
• C. $d=\frac{6}{\sqrt{14}}$ units
• D. $d=7$ units

Answer 1

Answer: (B), (C)

$d=\frac{6}{\sqrt{14}}$ units

Since dot product of dr's is equal to zero, so lines are mutually perpendicular.
$L_1,L_2$ and $L_3$ are mutually perpendicular lines and all these lines are passing through $P(1,2,3).$
$V=\frac{1}{6}\left|\left[\overrightarrow{PA}\overrightarrow{PB}\overrightarrow{PC}\right]\right|=\frac{1}{6}\times 2\times 3\times 6=6$


Also, $V=\frac{1}{3}\times \text{Base area}\times \text{Height}$
$\Rightarrow 6=\frac{1}{3}\times \sqrt{{(\frac{1}{2}\cdot 2\cdot 3)}^2+{(\frac{1}{2}\cdot 3\cdot 6)}^2+{(\frac{1}{2}\cdot 2\cdot 6)}^2}\times d$


$\therefore d=\frac{18}{\sqrt{9+81+36}}=\frac{18}{\sqrt{126}}=\frac{6}{\sqrt{14}}$ units

Alternate Solution :

Shifting origin at $P(1,2,3)$ and considering lines $L_1,L_2,L_3$ as $x-$axis, $y-$axis, $z-$axis respectively
Equation of plane $ABC$ is
$\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$
$\Rightarrow 3x+2y+z-6=0$

Perpendicular distance of the plane from the origin,
$d=\frac{6}{\sqrt{9+4+1}}=\frac{6}{\sqrt{14}}$