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

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., $4x+3y+5z=50$