Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Equation of plane passing through the intersection of the planes $2x-5y+z=3$ and $x+y+4z=5$ is

$(2x-5y+z-3)+\lambda (x+y+4z-5)=0$ $\Rightarrow (2+\lambda )\mathrm{x}+(-5+\lambda )\mathrm{y}+(1+4\lambda )-3-5\lambda =0$ .........(I)

which is parallel to the plane $x+3y+6z=1$.

Then $\frac{2+\lambda }{1}=\frac{-5+\lambda }{3}=\frac{1+4\lambda }{6}$

Then, $\frac{2+\lambda }{1}=\frac{-5+\lambda }{3}=\frac{1+4\lambda }{6}$

$\therefore \lambda =\frac{-11}{2}$

from eq. (i),

$-\frac{7}{2}x-\frac{21}{2}y-21z+\frac{49}{2}=0$

$\therefore x+3y+6z=7$

Hence, $\mathrm{k}=7$