Refer to the figure given below. The values of I_1 , I_2 and I_3...

Using the principle of superposition, we analyse the circuit with one source active at a time.

Case 1: Only the $10\text{ V}$ source is active (the $5\text{ V}$ source is replaced by a short circuit).

With the $5\text{ V}$ source shorted, the top and bottom nodes of the bridge are directly connected by a wire, forming a Wheatstone bridge.

Checking for balance:
Left arm ratio: $\dfrac{4}{4} = 1$
Right arm ratio: $\dfrac{2}{2} = 1$

The ratios are equal, so the bridge is balanced. Hence, no current flows through the middle branch connecting the two nodes, and the contribution to $I_2$ from the $10\text{ V}$ source is $0\text{ A}$.

Equivalent resistance of the balanced bridge:
$(4 + 2)\,\Omega \parallel (4 + 2)\,\Omega = 6\,\Omega \parallel 6\,\Omega = 3\,\Omega$

Total resistance including the series $1\,\Omega$ resistor:
$R_{eq} = 3 + 1 = 4\,\Omega$

Current drawn from the $10\text{ V}$ source:
$I_{1(10V)} = \dfrac{10}{4} = 2.5\text{ A}$

This current splits equally between the two $6\,\Omega$ branches, so the current through the bottom $4\,\Omega$ resistor is:
$I_{3(10V)} = \dfrac{2.5}{2} = 1.25\text{ A}$

Case 2: Only the $5\text{ V}$ source is active (the $10\text{ V}$ source is replaced by a short circuit).

Now the $5\text{ V}$ source is applied across the top and bottom nodes, and the $1\,\Omega$ resistor lies in the middle branch between the left and centre nodes.

For the left branch ($4\,\Omega$ and $4\,\Omega$ in series), the potential at the midpoint is $\dfrac{5}{2} = 2.5\text{ V}$.
For the right branch ($2\,\Omega$ and $2\,\Omega$ in series), the potential at the midpoint is also $\dfrac{5}{2} = 2.5\text{ V}$.

Since both midpoints are at the same potential, no current flows through the $1\,\Omega$ middle branch. Hence, the contribution to $I_1$ from the $5\text{ V}$ source is $0\text{ A}$.

Equivalent resistance seen by the $5\text{ V}$ source:
$(4 + 4)\,\Omega \parallel (2 + 2)\,\Omega = 8\,\Omega \parallel 4\,\Omega = \dfrac{8 \times 4}{8 + 4} = \dfrac{8}{3}\,\Omega$

Total current supplied by the $5\text{ V}$ source:
$I_{2(5V)} = \dfrac{5}{8/3} = \dfrac{15}{8} = 1.875\text{ A}$

Current through the left branch (which includes the bottom $4\,\Omega$ resistor):
$I_{3(5V)} = \dfrac{5}{8} = 0.625\text{ A}$

Applying superposition:
$I_1 = I_{1(10V)} + I_{1(5V)} = 2.5 + 0 = 2.5\text{ A}$
$I_2 = I_{2(10V)} + I_{2(5V)} = 0 + 1.875 = 1.875\text{ A}$
$I_3 = I_{3(10V)} + I_{3(5V)} = 1.25 + 0.625 = 1.875\text{ A}$

Hence, the correct option is $(1)\ I_1 = 2.5\text{ A},\ I_2 = 1.875\text{ A},\ I_3 = 1.875\text{ A}$.