Find the scalar equation of the plane that passes through the origin and the line of intersection of the planes 3x+4y-7z-2=0 and 2x+3y-4=0
Any plane that passes through the line of intersection of the planes 3x+4y-7z-2=0 and 2x+3y-4=0 can be written as:
$\displaystyle (3x+4y-7z-2)+\alpha(2x+3y-4)=0$ --------(1)
Since this plane passes through the origin,
$\displaystyle -2+\alpha(-4)=0$
$\displaystyle \alpha=-\frac{1}{2}$
Substituting in (1),
$\displaystyle (3x+4y-7z-2)-\frac{1}{2}(2x+3y-4)=0$
$\displaystyle 6x+8y-14z-4-2x-3y+4=0$
$\displaystyle 4x+5y-14z=0$