Question

The equation of a 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 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

The correct option is B $4x+3y+5z=50$

Given: $L_1:\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$(say), and $L_2:\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\mu$(say)
Let the points lies on $L_1,L_2$ are $P,Q$ respectively.
Hence
$P(3\lambda +1,\lambda +2,2\lambda +3)$
$Q(\mu +3,2\mu +1,3\mu +2)$
At the point of intersection
$3\lambda +1=\mu +3\Rightarrow 3\lambda -\mu =2\cdots \left(i\right)$
$\lambda +2=2\mu +1\Rightarrow \lambda -2\mu =-1\cdots \left(ii\right)$
Solving equation $\left(i\right),\left(ii\right)$
$\lambda =1$ and $\mu =1$
Hence, point of intersection is $P(4,3,5)$
Now plane passing through $(4,3,5)$ and at maximum distance from the origin must have directions of the normal as $4-0,3-0,5-0$.
$(x-4)4+(y-3)3+(z-5)5=0$
$4x+3y+5z=16+9+25$
$4x+3y+5z=50$