Question

Equation of the 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 has the largest distance from the

origin is :

• A. $7x+2y+4z=54$
• B. $3x+4y+5z=49$
• C. $4x+3y+5z=50$
• D. $5x+4y+3z=56$

Answer 1

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

$\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$ (say)

$\therefore x=3\lambda +1,y=\lambda +2,z=2\lambda +3$

$\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\mu$ (say)

.
$\therefore x=\mu +3,y=2\mu +1,z=3\mu +2$

$\Rightarrow 3\lambda +1=\mu +3$

$\Rightarrow \lambda +2=2\mu +1$

$\Rightarrow 2\lambda +3=3\mu +2$

So, $\lambda =1,\mu =1$

Point of intersection is $(4,3,5)$.

Equation of plane passing through $(4,3,5)$ such that the plane is at largest distance from origin will be possible, if the plane is perpendicular to line joining origin and $(4,3,5)$.

Hence, $(4,3,5)$ are the direction ratios of normal to the plane.
So, we can write the equation of plane as $4x+3y+5z=k$

Since it passes through $(4,3,5)$, we get the value of $k=50$
Hence, equation of plane is $4x+3y+5z=50$