If the solution of the differential equation $$(2x + 3y - 2)dx + (4x + 6y - 7)dy = 0$$, $$y(0) = 3$$, is $$\alpha x + \beta y + 3\log_e|2x + 3y - \gamma| = 6$$, then $$\alpha + 2\beta + 3\gamma$$ is equal to _______.

The given differential equation is$$(2x+3y-2)\,dx+(4x+6y-7)\,dy=0.$$

Rewrite it in the form $$\frac{dy}{dx}:$$
$$\frac{dy}{dx}= -\frac{2x+3y-2}{4x+6y-7}.$$

Notice that both numerator and denominator contain the linear expression $$2x+3y.$$ So set$$v=2x+3y.$$Then$$\frac{dv}{dx}=2+3\frac{dy}{dx}.$$

Substitute $$\frac{dy}{dx}=-\frac{v-2}{2v-7}$$ into the expression for $$\frac{dv}{dx}:$$
$$\frac{dv}{dx}=2+3\left(-\frac{v-2}{2v-7}\right)=\frac{v-8}{2v-7}.$$

This gives a separable equation:
$$\frac{2v-7}{v-8}\,dv=dx.$$

Integrate both sides. First write the left integrand as partial fractions:
$$\frac{2v-7}{v-8}=2+\frac{9}{v-8}.$$

Hence
$$\int\left(2+\frac{9}{v-8}\right)\,dv=\int dx,$$
$$2v+9\ln|v-8|=x+C.$$

Replace $$v$$ by $$2x+3y$$ and arrange terms:
$$4x+6y+9\ln|\,2x+3y-8\,|=x+C.$$

Bring $$x$$ to the left and divide by $$3$$ to match the required form:
$$x+2y+3\ln|\,2x+3y-8\,|=\frac{C}{3}.$$
Write $$\frac{C}{3}=K:$$
$$x+2y+3\log_e|\,2x+3y-8\,|=K.$$

Apply the initial condition $$y(0)=3$$:
$$0+2(3)+3\log_e|\,2(0)+3(3)-8\,|=K,$$
$$6+3\log_e|\,1\,|=K,\;\;K=6.$$

Thus the solution is
$$x+2y+3\log_e|\,2x+3y-8\,|=6.$$

Comparing with the required form
$$\alpha x+\beta y+3\log_e|\,2x+3y-\gamma\,|=6,$$
we identify $$\alpha=1,\quad\beta=2,\quad\gamma=8.$$

Finally,
$$\alpha+2\beta+3\gamma=1+2(2)+3(8)=1+4+24=29.$$

Hence the required value is $$29$$.