We know that $$P = {{{V^2}} \over R}$$

If potential is constant we have

$$P \propto {1 \over R}$$

Case I

Hence, this is a clear case of wheat stone bridge R$$_1$$ = 1 $$\Omega$$

Case II

Equivalent resistance (R$$_2$$)

$$\therefore$$ $${1 \over {{R_2}}} = {1 \over 2} + {1 \over 1} + {1 \over 2} = {{1 + 2 + 1} \over 2} = {4 \over 2} = 2\,\Omega $$

Hence, $${R_2} = {1 \over 2} = 0.5\,\Omega $$

It is clear that the equivalent resistance (R$$_2$$) will be less than 1 $$\Omega$$.

Case III

Hence, R$$_3$$ = 2 $$\Omega$$

Since, R$$_2$$ < R$$_1$$ < R$$_2$$

$$\therefore$$ P$$_2$$ > P$$_1$$ > P$$_3$$ ($$ \because $$ $$P \propto {1 \over R}$$)