Paragraph: Let S be the line of intersection of two planes x+y+z=1...

2 planes are $x+y+z=1$ and $2 x+3 y-4 z=8$. Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are $\mathrm{dr}^{\prime} \mathrm{s}$ is of lines of formed by intersection of planes then the lines is perpendicular to normal of both planes. $a+b+c=0$ ....(i) $\text{ and } 2 a+3 b-4 c=0$ .....(ii) $\frac{\mathrm{a}}{-4-3}=\frac{\mathrm{b}}{2+\mathrm{a}}=\frac{\mathrm{c}}{3-2}=\lambda$ (Solving (i) and (ii) $\mathrm{a}=-7 \lambda, \mathrm{ b}=6 \lambda, \mathrm{c}=\lambda$ Dr's or $\langle-7,6,1\rangle$.