Step 1:

$6x^2+5xy-6y^2+12x+5y+3=0$

Factorizing the 2nd degree terms,

$6x^2+5xy-6y^2=6x^2+9xy-4xy-6y^2$

$\qquad\qquad\qquad\;\;=3x(2x+3y)-2y(2x+3y)$

$\qquad\qquad\qquad\;\;=(3x-2y)(2x+3y)$

Step 2:

Therefore the equations of the asymptotes are of the form $3x-2y+l=0,2x+3y+m=0$

Then combined equation is $(3x-2y+l)(2x+3y+m)=0$----(1)

Step 3:

The combined equation of the asymptotes differs from that of the hyperbola only in the constant term.

Therefore their combined equation is of the form $6x^2+5xy-6y^2+12x+5y+3=0$------(2)

Step 4:

Eq(1) & Eq(2) represent the same pair of straight lines.

Comparing the coefficients of $x$ and $y$ we have,

$3m+2l=12$

$-2m+3l=5$

$\Delta=\begin{vmatrix}3 & 2\\-2 & 3\end{vmatrix}=9+4=13$

$\Delta_m=\begin{vmatrix}12 & 2\\5 & 3\end{vmatrix}=36-30=26$

$\Delta_l=\begin{vmatrix}3 & 12\\-2 & 5\end{vmatrix}=15+24=39$

Step 5:

$m=\large\frac{\Delta_m}{\Delta}=\large\frac{26}{13}$$=2$

$l=\large\frac{\Delta_l}{\Delta}=\large\frac{39}{13}$$=3$

The equations of the asymptotes are $3x-2y+3=0$,$2x+3y+2=0$