1. Given phase voltages

The instantaneous voltages are $V_X=V_0\sin\omega t$ $V_Y=V_0\sin\left(\omega t+\frac{2\pi}{3}\right)$ $V_Z=V_0\sin\left(\omega t+\frac{4\pi}{3}\right)$

These are three sinusoidal voltages of equal amplitude $V_0$, separated by phase difference $120^\circ$.

2. Voltage between terminals $X$ and $Y$

The instantaneous potential difference is $V_{XY}=V_X-V_Y$ So, $V_{XY}=V_0\sin\omega t - V_0\sin\left(\omega t+\frac{2\pi}{3}\right)$

Use the identity $\sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

Let $A=\omega t,\qquad B=\omega t+\frac{2\pi}{3}$ Then $V_{XY}=2V_0\cos\left(\omega t+\frac{\pi}{3}\right)\sin\left(-\frac{\pi}{3}\right)$ Since $\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}$ we get $V_{XY}=-\sqrt{3}V_0\cos\left(\omega t+\frac{\pi}{3}\right)$

Thus the amplitude of $V_{XY}$ is $V_{XY}^{\text{max}}=\sqrt{3}V_0$ Hence its rms value is $V_{XY}^{\text{rms}}=\frac{\sqrt{3}V_0}{\sqrt{2}}=V_0\sqrt{\frac{3}{2}}$

So Option A is correct.

3. Voltage between terminals $Y$ and $Z$

Similarly, $V_{YZ}=V_Y-V_Z$ Since $V_Y$ and $V_Z$ also have the same amplitude and phase difference $120^\circ$, the result must be the same magnitude.

Let us verify: $V_{YZ}=V_0\sin\left(\omega t+\frac{2\pi}{3}\right)-V_0\sin\left(\omega t+\frac{4\pi}{3}\right)$ Again, the difference of two equal-amplitude sinusoids separated by $120^\circ$ has amplitude $\sqrt{3}V_0$ Therefore, $V_{YZ}^{\text{rms}}=\frac{\sqrt{3}V_0}{\sqrt{2}}=V_0\sqrt{\frac{3}{2}}$

So Option B is incorrect.

4. Check Option C

Option C says $V_{XY}^{\text{rms}}=V_0$ But we found $V_{XY}^{\text{rms}}=V_0\sqrt{\frac{3}{2}}$ Hence Option C is incorrect.

5. Check dependence on terminal choice

The three voltages form a balanced three-phase system. The rms line voltage between any two terminals is the same: $V_{XY}^{\text{rms}}=V_{YZ}^{\text{rms}}=V_{ZX}^{\text{rms}}=V_0\sqrt{\frac{3}{2}}$

Therefore the reading is independent of the choice of the two terminals.

So Option D is correct.

6. Final answer

Correct options are: $\boxed{A, D}$

This matches the stored correct answer.