simply quadratic!!!!!

By Vieta's formula, product of the roots is 2m/3 $=6$ which gives us $m=9$ on cross-multiplication.

Since $2$ and $3$ are the roots, we have

$(x-2)(x-3)=0$

$x^2-3x-2x+6=0$

$x^2-5x+6=0$

Multiplying it by $3$ , gives

$3x^2-15x+18=0$

So,

$2m=18$

It follows that,

$\boxed{m=9}$

If $2,3$ are the roots then we can put $x=2,3$ and the equation will be equal to $0$ .We get 2 equations: $\color{#3D99F6}{3(2)^2-2k(2)+2m=0\rightarrow(after\;simplification) 2k-m=6}$ $\color{#3D99F6}{3(3)^2-2k(3)+2m=0\rightarrow(after\;simplification) 6k-2m=27}$ $m=2k-6$ .Substituting this into the second equation and simplifying,we get $k=7.5$ .Substituting this into the first equation gives us $\boxed{m=9}$

subtract both the equations to get the value of k=7.5...
then substitute 'k' in both equations to get M=9.