Question

Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ and twice of its y- intercept is three times its z-intercept.

Answer 1

Given:$x+y+z=1$

$2x+3y+4z=5$

Required equation of plane is $x+y-1+z+\lambda (2x+3y+4z-5)=0$ for some $\lambda$

$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z=1+5\lambda$

According to the question,

$2\left(\frac{1+5\lambda }{1+3\lambda }\right)=3\left(\frac{1+5\lambda }{1+4\lambda }\right)$

$\Rightarrow 2(1+5\lambda )(1+4\lambda )=3(1+5\lambda )(1+3\lambda )$

$\Rightarrow 2(1+4\lambda )=3(1+3\lambda )$

$\Rightarrow 2+8\lambda =3+9\lambda$

$\Rightarrow 9\lambda -8\lambda =-3+2$

$\Rightarrow 9\lambda -8\lambda =-1$

$\Rightarrow \lambda =-1$

Thus, the equation of required plane is

$-x-2y-3z=-4$

or $x+2y+3z=4$