Point A is the intersection of y = x + $$\alpha$$ and y = 2 $$-$$ x lines.

$$\therefore$$ $$y = 2 - y + \alpha $$

$$ \Rightarrow 2y = 2 + \alpha $$

$$ \Rightarrow y = {{2 + \alpha } \over 2}$$

and $$x = 2 - {{2 + \alpha } \over 2} = {{2 - \alpha } \over 2}$$

$$\therefore$$ Point $$A = \left( {{{2 - \alpha } \over 2},{{2 + \alpha } \over 2}} \right)$$

$$\therefore$$ AN = y-coordinate of point $$A = {{2 + \alpha } \over 2}$$

Point B is the intersection of $$y = 2 + x$$ and $$y = - x - \alpha $$ lines.

$$\therefore$$ $$y = 2 - y - \alpha $$

$$ \Rightarrow 2y = 2 - \alpha $$

$$ \Rightarrow y = {{2 - \alpha } \over 2}$$

$$\therefore$$ $$x = {{2 - \alpha } \over 2} - 2 = {{2 - \alpha - 4} \over 2} = {{ - \alpha - 2} \over 2}$$

$$\therefore$$ Point $$B = \left( {{{ - \alpha - 2} \over 2},{{2 - \alpha } \over 2}} \right)$$

$$\therefore$$ $$BM = y - $$coordinate of point $$B = {{2 - \alpha } \over 2}$$

Area of the common region $$BRAE$$

$$ = \Delta CDE - \left( {\Delta BCR + \Delta ARD} \right)$$

$$ = {1 \over 2} \times 4 \times 2 - \left( {{1 \over 2}( - \alpha + 2) \times BM + {1 \over 2} \times (2 + \alpha ) \times AN} \right)$$

$$ = 4 - \left( {{1 \over 2} \times (2 - \alpha ) \times {{(2 - \alpha )} \over 2} + {1 \over 2} \times (2 + \alpha ) \times {{2 + \alpha } \over 2}} \right)$$

$$ = 4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right]$$

Given, $$4 - \left[ {{{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4}} \right] = {3 \over 2}$$

$$ \Rightarrow {{{{(2 - \alpha )}^2}} \over 4} + {{{{(2 + \alpha )}^2}} \over 4} = {5 \over 2}$$

$$ \Rightarrow {(2 - \alpha )^2} + {(2 + \alpha )^2} = 10$$

$$ \Rightarrow 4 + {\alpha ^2} - 4\alpha + 4 + {\alpha ^2} + 4\alpha = 10$$

$$ \Rightarrow 2{\alpha ^2} + 8 = 10$$

$$ \Rightarrow 2{\alpha ^2} = 2$$

$$ \Rightarrow {\alpha ^2} = 1$$

$$ \Rightarrow \alpha = \, \pm \,1$$

Given that $$\alpha > 0$$ so accepted value of $$\alpha = + \,1$$.

Now, $$0 \le y \le x + 2\alpha $$ and $$|x| \le 1$$

$$ \Rightarrow 0 \le y \le x + 2$$ and $$ - 1 \le x \le 1$$

Area of $$ABCD = {1 \over 2}(1 + 3) \times (1 - ( - 1))$$

$$ = {1 \over 2} \times 4 \times 2$$

$$ = 4$$ sq. unit