Paragraph: Let L: x+y+z+4=0=2 x-y-z+8 be a line and P : x+2 y+3...

We have $\mathrm{L}: \mathrm{x}+\mathrm{y}+\mathrm{z}+4=0=2 \mathrm{x}-\mathrm{y}-\mathrm{z}-8$ and P: x+2 y+3 z+1=0 is a plane. Let Dr's of line $\langle$ a, b, c $\rangle$ Since, line $L$ obtained by intersection of planes $x+y+z+4=0 \text{ and } 2 x-y-z-8=0$ So, $\mathrm{a}+\mathrm{b}+\mathrm{c}=0$ .....(i) $2 a-b-c=0$ ....(ii) (line is perpendicular to normal of both plane) $$\begin{aligned} & \frac{\mathrm{a}}{-1+1}=\frac{\mathrm{b}}{2+1}=\frac{\mathrm{c}}{-1-2}=\lambda \\ & \Rightarrow \mathrm{a}=0, \mathrm{ b}=3 \lambda, \mathrm{c}=-3 \lambda \end{aligned}$$ (solving (i) and (ii) Dr's of line is $\langle 0,3,-3\rangle or \langle 0,1,-1\rangle$