Consider three equidistant slits A, B, and C (with B located...

Let the distance between adjacent slits be $d$. The path length from slit A to point $P_0$ is $x_A = D$.

The path length from slit B to $P_0$ is $x_B = \sqrt{D^2 + d^2}$.
For $D \gg d$, the path difference is:
$\Delta x_B = x_B - x_A = \sqrt{D^2 + d^2} - D \approx \dfrac{d^2}{2D}$
We are given that $\Delta x_B = \dfrac{\lambda}{3}$, which implies the phase difference is $\phi_B = \dfrac{2\pi}{\lambda} \Delta x_B = \dfrac{2\pi}{3}$.

The path length from slit C to $P_0$ is $x_C = \sqrt{D^2 + (2d)^2}$.
The path difference is:
$\Delta x_C = x_C - x_A \approx \dfrac{4d^2}{2D} = 4 \Delta x_B = \dfrac{4\lambda}{3}$
The phase difference for the wave from C is $\phi_C = \dfrac{2\pi}{\lambda} \Delta x_C = \dfrac{8\pi}{3}$.

Let the amplitude of the wave from each slit be $A_0$, so $I_0 = A_0^2$. The resultant complex amplitude at $P_0$ is:
$A_{net} = A_0 + A_0 e^{i\phi_B} + A_0 e^{i\phi_C}$
$A_{net} = A_0 \left( 1 + e^{i 2\pi/3} + e^{i 8\pi/3} \right)$

Since $e^{i 8\pi/3} = e^{i (2\pi + 2\pi/3)} = e^{i 2\pi/3}$, we have:
$A_{net} = A_0 \left( 1 + 2 e^{i 2\pi/3} \right)$
$A_{net} = A_0 \left( 1 + 2 \left( -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \right) \right) = A_0 (i\sqrt{3})$

The resultant intensity at $P_0$ is:
$I = |A_{net}|^2 = (\sqrt{3} A_0)^2 = 3 A_0^2 = 3 I_0$