Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

We have, line of intersection of the planes
$x+2y+3z=4$and$4x+3y+2z=1$
$\therefore$Equation of plane passing through the given planes is
$\left(x+2y+3z-4\right)+\lambda \left(4x+3y+2z-1\right)=0$
$\Rightarrow \left(1+4\lambda \right)x+\left(2+3\lambda \right)y+\left(3+2\lambda \right)+\left(-4-\lambda \right)=0$
Since, plane passing through origin
$\therefore -4-\lambda =0\Rightarrow \lambda =-4$
Now, equation of plane is
$\left(1-16\right)x+\left(2-12\right)y+\left(3-8\right)z+0=0$
$\Rightarrow -15x-10y-5z=0$
$\Rightarrow 3x+2y+z=0$
$\therefore$Direction rations of the normal to the plane are 3, 2, 1.