Given,

The lines $L_1\equiv 2x+y+3=0, L_2\equiv kx+2y-3=0$ and $L_3\equiv 3x-2y+1=0$ are concurrent,

So by using the condition of concurrent we get,

$\left|\begin{array}{ccc}2& 1& 3\\ k& 2& -3\\ 3& -2& 1\end{array}\right|=0$

$\Rightarrow 2\left(2-6\right)-1\left(k+9\right)+3\left(-2k-6\right)=0$

$\Rightarrow k=-5$

Now $L_2=-5x+2y-3=0$

So, slope of $L_2$ will be $m_1=\frac{5}{2}$

And slope of line $2x-5y+7=0$ will be $m_2=\frac{2}{5}$

Now we know that $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow \cos \theta =\left|\frac{1+m_1{m}_2}{\sqrt{{\left(m_1-m_2\right)}^2+{\left(1+m_1{m}_2\right)}^2}}\right|$

$\Rightarrow \cos \theta =\left|\frac{1+{\displaystyle 1}}{\sqrt{{\left({\displaystyle \frac{5}{2}}-{\displaystyle \frac{2}{5}}\right)}^2+{\left(1+1\right)}^2}}\right|$

$\Rightarrow \cos \theta =\left|\frac{2}{\sqrt{{\displaystyle \frac{841}{100}}}}\right|=\frac{20}{29}$