JEE Main 2014 — Three Dimensional Geometry Question with Solution

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$L_1:\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$

$L_2:\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\mu$

$\begin{array}{ccc}\Rightarrow x=3\lambda +1& \text{and}& x=\mu +3\\ y=\lambda +2& & y=2\mu +1\\ z=2\lambda +3& & z=3\mu +2\end{array}$

$\begin{array}{cccc}\Rightarrow & 3\lambda +1=\mu +3& \Rightarrow & 3\lambda -\mu =2\end{array}$

$\begin{array}{ccc}\lambda +2=2\mu +1& \Rightarrow & \lambda -2\mu =-1\end{array}$

$\begin{array}{ccc}2\lambda +3=3\mu +2& \Rightarrow & 2\lambda -3\mu =-1\end{array}$

Solving $\mu =1, \lambda =1$

$\therefore$  Point of intersection $P\left(4,3,5\right)$ and D.R. of normal is (largest distance is normal distance which is $OP$) $\left(4,3,5\right)$.

Equation of plane $4x+3y+5z+d=0$ this passes through the point  $\left(4,3,5\right)$

$\Rightarrow 4\times 4+3\times 3+5\times 5+d=0$

$\Rightarrow d=-50$

So, equation of plane is $4x+3y+5z-50=0$