Question

The equation of the plane which is equidistant from the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$ is $Ax+By+Cz-9=0$ then the value of $A+B+C=$

• A. $x+5y-6z=0$
• B. $x+7y-5z-9=0$
• C. $3x+5y-7z=10$
• D. $3x+8y-9z=6$

Answer 1

The correct option is B $x+7y-5z-9=0$

Equation of plane which contains the line $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$ is

$(x-2-3(y-3))+\lambda (2(y-3)-(z-1))=0$

$\Rightarrow x+(2\lambda -3)y-\lambda z+7-5\lambda =0$

If the plane is parallel to $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$

$\Rightarrow 1+(2\lambda -3)2-3\lambda =0\Rightarrow \lambda =5$

So equation of plane which contains second line and parallel to first is $x+7y-5z-18=0.$

Similarly, equation of plane which contains first line and parallel to second is $(x-1)+7(y-2)-5(z-3)=0$

$\Rightarrow x+7y-5z=0.$

Hence equidistant plane is $\frac{(x+7y-5z-18)+(x+7y-5z)}{2}=0$

$\Rightarrow x+7y-5z-9=0$