If the solution of the differential equation ( 2 x + 3 y - 2 ) d x + ( 4 x + 6 y - 7 ) d y = 0 , y ( 0 ) = 3…

Given: $(2x+3y-2)dx+(4x+6y-7)dy=0$

$\Rightarrow \frac{dy}{dx}=\frac{-\left(2x+3y-2\right)}{\left(4x+6y-7\right)} ...\left(i\right)$

Let, $2x+3y-2=t$

$\Rightarrow 2+3\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}} ...\left(ii\right)$

Also, $4x+6y-4=2t$

$\Rightarrow 4\mathrm{x}+6\mathrm{y}-7=2\mathrm{t}-3 ...\left(iii\right)$

Using $\left(i\right), \left(ii\right) \text{and} \left(iii\right)$,

$\Rightarrow \frac{1}{3}\left(\frac{dt}{dx}-2\right)=\frac{-t}{2t-3}$

$\Rightarrow \frac{dt}{dx}-2=\frac{-3t}{2t-3}$

$\Rightarrow \frac{dt}{dx}=\frac{-3t}{2t-3}+2$

$\Rightarrow \frac{dt}{dx}=\frac{-3t+4t-6}{2t-3}$

$\Rightarrow \frac{dt}{dx}=\frac{t-6}{2t-3}$

$\Rightarrow \int \frac{2t-3}{t-6}dt=\int dx$

$\Rightarrow \int \left(\frac{2t-12}{t-6}+\frac{9}{t-6}\right)·dt=x+c$

$\Rightarrow 2t+9\log (t-6)=x+c$

$\Rightarrow 2(2x+3y-2)+9\log (2x+3y-8)=x+c ...\left(iv\right)$

It is given that $y\left(0\right)=3$.

$\Rightarrow 2(0+9-2)+9\log (0+9-8)=0+c$

$\Rightarrow 14+9\log \left(1\right)=c$

$\Rightarrow c=14$

Using $\left(iv\right)$,

$\Rightarrow 4x+6y-4+9\log (2x+3y-8)=x+14$

$\Rightarrow 3x+6y+9\log (2x+3y-8)=18$

$\Rightarrow x+2y+3\log (2x+3y-8)=6$

Comparing with $\alpha x+\beta y+3{\log }_e|2x+3y-\gamma |=6$.

$\Rightarrow \alpha =1, \beta =2, \gamma =8$

$\Rightarrow \alpha +2\beta +3\gamma =1+4+24$

$\Rightarrow \alpha +2\beta +3\gamma =29$