Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane passing through the intersection of the planes $x+y+z-1=0$ and $2x+3y-z+4=0$ is given by:

$(x+y+z-1)+\lambda (2x+3y-z+4)=0$

$(1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+(4\lambda -1)=0$

Since the required plane is parallel to the $Y$-axis, its normal is perpendicular to the $Y$-axis. The direction ratios of the $Y$-axis are $(0,1,0)$.

Taking the dot product of the normal vector and the direction vector of the $Y$-axis as zero:

$0(1+2\lambda )+1(1+3\lambda )+0(1-\lambda )=0$

$1+3\lambda =0\Rightarrow \lambda =-\frac{1}{3}$

Substituting $\lambda =-\frac{1}{3}$ in the equation of the plane, we get:

$\left(1-\frac{2}{3}\right)x+\left(1-1\right)y+\left(1+\frac{1}{3}\right)z+\left(-\frac{4}{3}-1\right)=0$

$\frac{1}{3}x+\frac{4}{3}z-\frac{7}{3}=0$

$x+4z-7=0$

Answer: $x+4z-7=0$