1. Understand the arrangement

   The capacitor gap is filled with three dielectric slabs of equal thickness placed one after another along the separation between plates.

   Hence they behave like three capacitors in series.

2. Capacitance of each slab

   Let the plate area be $A$ and total separation be $d$.

   Since the three slabs have equal thickness, each slab has thickness:

   $$d_1=d_2=d_3=\frac{d}{3}$$

   For a dielectric slab of dielectric constant $K_i$ and thickness $d/3$,

   $$C_i=\frac{\varepsilon_0 K_i A}{d/3}=\frac{3\varepsilon_0 K_i A}{d}$$

3. Series combination

   For capacitors in series,

   $$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$

   Substituting:

   $$\frac{1}{C}=\frac{d}{3\varepsilon_0 A}\left(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}\right)$$

   With $K_1=10, K_2=12, K_3=14$,

   $$\frac{1}{C}=\frac{d}{3\varepsilon_0 A}\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}\right)$$

4. Equivalent dielectric constant

   If a single dielectric of constant $K$ fully fills the capacitor, then

   $$C=\frac{\varepsilon_0 K A}{d}$$

   Therefore,

   $$\frac{d}{\varepsilon_0 K A}=\frac{d}{3\varepsilon_0 A}\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}\right)$$

   Cancelling common factors:

   $$\frac{1}{K}=\frac{1}{3}\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}\right)$$

5. Calculate the value

   First,

   $$\frac{1}{10}+\frac{1}{12}+\frac{1}{14} =\frac{42+35+30}{420} =\frac{107}{420}$$

   So,

   $$\frac{1}{K}=\frac{107}{1260}$$ $$K=\frac{1260}{107}\approx 11.78$$

   Closest option:

   $$K\approx 12$$

6. Check options

   • A: 12 ✓
   • B: 36 ✗
   • C: 14 ✗
   • D: 4 ✗

So the correct answer is Option A: 12.