If the angle between the lines represented by $2x^2+5xy+3y^2+6x+7y+4=0$ is ${\tan }^{-1}\left(m\right)$ and $a^2+b^2-ab-a-b+1\le 0$, then $2a+3b=$

$a=2,2h=5,b=3,2g=6,2f=7,c=4$

$\tan \theta =\frac{2\sqrt{h^2-ab}}{|a+b|}=\frac{1}{5}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{{\displaystyle 1}}{{\displaystyle 5}}\right),m=\frac{1}{5}$

$2a^2+2b^2-2ab-2a-2b+2\le 0$

$\Rightarrow a^2-2a\left(1\right)+1^2+b^2-2b\left(1\right)+1^2+a^2$

$-2ab+b^2\le 0$

$\Rightarrow {(a-1)}^2+{(b-1)}^2+{(a-b)}^2\le 0$

which is possible only if $a-1=0,b-1=0,a-b=0\Rightarrow a=b=1$