MHTCET 16 May 2024 Shift 1 Paper

Concept:Any plane passing through the line of intersection of two given planes can be written as a linear combination of their equations.For parallelism to the $Y$-axis, the normal vector of the plane must be perpendicular to the $Y$-axis, so the coefficient of $y$ must be zero.Explanation:The equation of the plane through the intersection of$x+y+z-1=0$ and $2x+3y-z+4=0$ is$$(x+y+z-1)+\lambda(2x+3y-z+4)=0.$$Expanding, we get$$(1+2\lambda)x+(1+3\lambda)y+(1-\lambda)z+(4\lambda-1)=0.$$Since this plane is parallel to the $Y$-axis, its normal vector is perpendicular to $(0,1,0)$.Therefore, the coefficient of $y$ must be zero:$$1+3\lambda=0 \Rightarrow \lambda=-\frac{1}{3}.$$Substitute $\lambda=-\frac{1}{3}$:Coefficient of $x$: $1+2\left(-\frac13\right)=\frac13$Coefficient of $z$: $1-\left(-\frac13\right)=\frac43$Constant term: $4\left(-\frac13\right)-1=-\frac73$Multiplying the equation by $3$ gives:$$x+4z-7=0.$$Answer:The required equation of the plane is $x+4z-7=0$.Hence, the correct option is B.