Question

Equation of plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at greatest distance from the point $(0,0,0)$ is.

• A. $4 x+3 y+5 z=25$
• B. $4 x+3 y+5 z=50$
• C. $3 x+4 y+5 z=49$
• D. $x+7 y-5 z=2$

Answer 1

Answer: (B)

A general point on the line
$L _{1}: \frac{ x -1}{3}=\frac{ y -2}{1}=\frac{ z -3}{2}=\lambda...$(i)
will be $(3 \lambda+1, \lambda+2,2 \lambda+3)$ and similarly on
$L _{2}: \frac{ x -3}{1}=\frac{ y -1}{2}=\frac{ x -2}{3}=\mu...$(ii)
will be $(\mu+3,2 \mu+1,3 \mu+2)$
From(i) and (ii) we get $\lambda=\mu=1$
and point of intersection is $P (4,3,5)$.
The plane passing through $P (4,3,5)$
which is at maximum distance from origin will have normal along $\overrightarrow{ OP }$
i.e., $\overrightarrow{ n }=4 \hat{ i }+3 \hat{ j }+5 \hat{ k }$
Hence the plane will be
$4(x-4)+3(y-3)+5(z-5)=0$
i.e., $4 x+3 y+5 z=50$