Question

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

Answer 1

Given: $L_1:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $L_2:\frac{x-2}{3}=\frac{y-3}{1}=\frac{z-1}{2}$
Let DR's of plane be $a,b,c$
The equation of the plane containing line $L_1$ is
$a(x-1)+b(y-2)+c(z-3)=0\cdots \left(i\right)$
And also product of their DR's will be zero.
$a+2b+3c=0\cdots \left(ii\right)$
And product of DR of plane and line $L_2$ will also be zero.
$\therefore 3a+b+2c=0\cdots \left(iii\right)$
Solving $\left(ii\right),\left(iii\right)$
$\frac{a}{1}=\frac{b}{7}=\frac{c}{-5}=k$
putting in equation $\left(i\right)$
$k(x-1)+7k(y-2)-5k(z-3)=0$
$\Rightarrow x+7y-5z=0$
$\therefore A+B+C=1+7-5=3$