If  ${\left(1+2\mathrm{x}+3{\mathrm{x}}^2\right)}^{10}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+...+{\mathrm{a}}_{20}{\mathrm{x}}^{20}$ then

The given expansion is ${\left(1+2x+3x^2\right)}^{10}={\left[1+x\left(2+3x\right)\right]}^{10}$

By expanding the above expression we get 

                  ${\left[1+\mathrm{x}(2+3\mathrm{x})\right]}^{10}=1+10{\mathrm{c}}_1\mathrm{x}(2+3\mathrm{x})+\cdots +10{\mathrm{c}}_{10}{\mathrm{x}}^{10}(2+3\mathrm{x}{)}^{10}$

The coefficient of $x$ is $a_1=10\left(2\right)=\boxed{20}$

The coefficient of $x^2$ is $a_2=$$10\left(3\right)+\frac{10·9}{2}\left(4\right)=30+180=\boxed{210}$

The coefficient of $x^3$ is 

             $\begin{array}{rcl}{a}_4& =& \frac{10·9}{2}\left(9\right)+\frac{10·9·8}{6}\left(36\right)+\frac{10·9·8·7}{24}\left(16\right)\\ & =& 405+4320+3360\\ & =& \boxed{8085}\end{array}$

 The value of $a_6$ must be more than 8085, its value does not equal to 12.