1. Set up the Wheatstone bridge condition

   Four resistances are arranged in cyclic order:

   $15\,\Omega,\;12\,\Omega,\;4\,\Omega,\;10\,\Omega$

   Let the resistance to be connected in parallel with the $10\,\Omega$ resistor be $R$.

   Then the effective resistance of that branch becomes

   $R_{eq}=\frac{10R}{10+R}$

2. Use the balance condition of Wheatstone bridge

   For a balanced Wheatstone network, the product of opposite arms must be equal:

   $15 \times 4 = 12 \times R_{eq}$

   Substitute $R_{eq}$:

   $15 \times 4 = 12 \times \frac{10R}{10+R}$

   $60 = \frac{120R}{10+R}$

3. Solve for $R$

   $60(10+R)=120R$

   $600+60R=120R$

   $600=60R$

   $R=10\,\Omega$

4. Verification

   If $R=10\,\Omega$, then parallel combination with $10\,\Omega$ is

   $R_{eq}=\frac{10\times 10}{10+10}=5\,\Omega$

   Now check balance:

   $15\times 4 = 60, \qquad 12\times 5 = 60$

   Both are equal, so the bridge is balanced.

Hence, the required resistance is:

$\boxed{10\,\Omega}$