Solution

Comparing $$2x+3y =7$$ with $$a_1x+b_1y+c_1=0$$,
we get,
$$a_1=2, b_1=3,c_1=-7$$

Comparing $$\left ( a-b \right )x + \left ( a +b \right )y = 3a + b - 2$$ with $$a_2x+b_2y+c_2=0$$,
we get,
$$a_2=\left ( a-b \right ), b_2=\left ( a +b \right ), c_2=-(3a + b - 2)$$

Given system of equation have infinite solutions:
$$\Rightarrow \dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$$

$$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$$

From $$\dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$$

$$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}$$

$$\Rightarrow 2a + 2b = 3a-3b$$

$$\Rightarrow a-5b = 0 ......... (I)$$

Now from $$\dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$$

$$\Rightarrow \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$$

$$\Rightarrow9a + 3b -6 = 7a + 7b$$

$$\Rightarrow 9a-7a+3b-7b =6$$

$$\Rightarrow 2a-4b = 6 ...... (II)$$

Now, for solving eq $$I$$ and eq $$II$$

$$a-5b = 0 $$

$$2a-4b = 6$$

Multiplying eq $$I$$ by $$2$$

$$2a - 10b =0$$

$$2a - 4b =6$$

on subtraction we get

$$\Rightarrow -6b = -6$$

$$\Rightarrow b = 1$$

Now, substituting $$b$$ value in eq $$I$$

$$\Rightarrow 2a-10\left ( 1 \right ) = 0$$

$$2a - 10 = 0$$

$$ 2a = 10$$

$$ a = \dfrac{10}{2}$$

$$ a = 5$$

$$\Rightarrow a = 5 , b = 1$$