The general term in the expansion of
$\left(2-x+3 x^{2}\right)^{6}$ is
$\frac{6 !}{r ! s ! t !} 2^{r}(-x)^{s}\left(3 x^{2}\right)^{t}$, where $r+s+t=6$
$=\frac{6 !}{r ! s ! t !} 2^{r} \times(-1)^{s} \times 3^{t} \times x^{s+2 t}$
where $r+s+t=6$
For the coefficient of $x^{5}$, we must have
$s+2 t=5$
But $r+s+t=6$
$\therefore s=5-2 t$ and $r=1+t$
where, $0 \leq r, s, t \leq 6$
Now, $t=0 \Rightarrow r=1, s=5$
$t=1 \Rightarrow r=2, s=3$
$t=2 \Rightarrow r=3, s=1$
Thus, there are three terms containing $x^{5}$ and coefficient of $x^{5}$
$=\frac{6 !}{1 ! 5 ! 0 !} \times 2^{1} \times(-1)^{5} \times 3+\frac{6 !}{2 ! 3 ! 1 !} \times 2^{2}$
$\times(-1)^{3} \times 3^{1}+\frac{6 !}{3 ! 1 ! 2 !} \times 2^{3} \times(-1)^{1} \times 3^{2}$
$=-12-720-4320$
$=-5052$