MHTCET 11 Apr 2026 Shift 2 Paper

Concept:The matrix equation becomes a system of linear equations. Solve for $a,b,c$, then use the point-to-plane distance formula.Explanation:Expanding the matrix product gives:$$2a+b-c=4$$$$-a+2b+c=0$$$$a-b+2c=-2$$From the second equation, $c=a-2b$.Substitute into the first equation:$2a+b-(a-2b)=4$, so $a+3b=4$.Substitute into the third equation:$a-b+2(a-2b)=-2$, so $3a-5b=-2$.Solving these gives $a=1$ and $b=1$.Thus $c=1-2(1)=-1$.So the point is $P(1,1,-1)$.Distance from $(x_1,y_1,z_1)$ to plane $Ax+By+Cz+D=0$ is:$$\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$$The given plane is $2x+y+2z=10$, i.e.$$2x+y+2z-10=0$$So $A=2$, $B=1$, $C=2$, $D=-10$.Substitute $P(1,1,-1)$:$$\text{Distance}=\frac{|2(1)+1(1)+2(-1)-10|}{\sqrt{2^2+1^2+2^2}}$$$$=\frac{|2+1-2-10|}{\sqrt{4+1+4}}=\frac{9}{3}=3$$Answer:The distance is $3$ units. Hence the correct option is B.