Question

The equation of the line which intersect each of the two lines $2x+y-1=0=x-2y+3z$ and $3x-y+z+2=0=4x+5y-2z-3=0$ and is parallel to $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is

• A. $4x+7y-6z-1=0=2x-7y+4z+3$
• B. $4x+7y-6z-4=0=2x-7y+4z+2$
• C. $4x+7y-6z-3=0=2x-7y+4z+7$
• D. $4x+7y-6z+7=0=2x-7y+4z-3$

Answer 1

Answer: (C)

Equation of the first plane is $\left(2x+y-1\right)+\lambda \left(x-2y+3z\right)=0$
$\Rightarrow x\left(2+\lambda \right)+y\left(1-2\lambda \right)+3\lambda z-1=0$
Equation of the second plane is $\left(3x-y+z+2\right)+u\left(4x+5y-2z-3\right)=0$
$\Rightarrow x\left(3+4u\right)+y\left(-1+5u\right)+z\left(1-2u\right)+2-3u=0$
Required line must lie in both planes and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
$\Rightarrow \left(2+\lambda \right)1+\left(1-2\lambda \right)2+\left(3\lambda \right)3=0\Rightarrow \lambda =-\frac{2}{3}$
$\Rightarrow \left(3+4u\right)1+\left(-1+5u\right)2+\left(1-2u\right)\left(3\right)=0$
$\Rightarrow 8u+4=0\Rightarrow u=-\frac{1}{2}$
Equation of both planes is $4x+7y-6z-3=0$ and $2x-7y+4z+7=0$
So the equation of the required line is
$4x+7y-6z-3=0=2x-7y+4z+7$