To solve this problem, let us determine the speed of the river step by step:

---

Variables:

1. Speed of Pooja in still water = $v_p$ km/min.
2. Speed of river current = $v_r$ km/min.
3. Time Pooja took to swim downstream back to the bank = $T_d$ minutes.
4. Distance Pooja swam upstream before realizing she lost her cap = 3 km.
5. Time taken by Pooja to realize = 12 minutes.

---

Step 1: Speed of the cap

• The cap was stationary relative to the river and moved downstream due to the current.
• In 12 minutes, the cap traveled a distance of: $d = 12 \cdot v_r \quad \text{(as speed = distance/time)}.$

---

Step 2: Time for Pooja to return

• Pooja started swimming back to the bank after realizing the cap was missing.
• She swam downstream for $T_d$ minutes such that she and the cap reached the bank at the same time.

---

Step 3: Relative speed and equations

Downstream speed of Pooja:

$v_{d} = v_p + v_r \quad \text{(Pooja's speed downstream)}.$

Upstream distance covered by the cap in $T_d + 12$:

• The cap traveled the full upstream distance of 3 km in $12 + T_d$ minutes: $3 = v_r \cdot (12 + T_d).$

Distance Pooja covered downstream:

• Pooja swam back the same 3 km downstream in $T_d$ minutes: $3 = (v_p + v_r) \cdot T_d.$

---

Step 4: Solve equations

From the equations:

1. $T_d = \frac{3}{v_p + v_r}$.
2. Substituting $T_d$ in $3 = v_r \cdot (12 + T_d)$: $3 = v_r \cdot \left(12 + \frac{3}{v_p + v_r}\right).$

Simplify: $3 = 12v_r + \frac{3v_r}{v_p + v_r}.$

Multiply through by $v_p + v_r$ to eliminate the fraction: $3(v_p + v_r) = 12v_r(v_p + v_r) + 3v_r.$

Expand and rearrange terms: $3v_p + 3v_r = 12v_r v_p + 12v_r^2 + 3v_r.$

Simplify further: $3v_p = 12v_r v_p + 12v_r^2.$

Divide through by $3$: $v_p = 4v_r v_p + 4v_r^2.$

Rearrange: $v_p (1 - 4v_r) = 4v_r^2.$

Solve for $v_r$: This is a quadratic equation in $v_r$. Use specific $v_p$ (Pooja’s swimming speed) to calculate $v_r$.

---

If you share Pooja's swimming speed, I can compute the numerical solution for $v_r$.