This problem combines concepts from matrices (trace, adjugate) and trigonometry. The key is to carefully use the given conditions to evaluate the truthfulness of the statements. We will first establish crucial relationships between $x, y, z$ using the trigonometric condition.

1. Key Concepts and Formulas

• Diagonal Matrix: A diagonal matrix $R = \text{diag}(x, y, z)$ has entries only on its main diagonal: $R=\left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right)$
• Trace of a Matrix: For a square matrix $M$, $\text{Trace}(M)$ is the sum of its diagonal entries. For $R$, $\text{Trace}(R) = x+y+z$.
• Determinant of a Diagonal Matrix: For $R = \text{diag}(x, y, z)$, its determinant is $\det(R) = xyz$.
• Adjugate of a Matrix Property: For any invertible square matrix $M$ of order $n$, $\operatorname{adj}(\operatorname{adj}(M)) = (\det M)^{n-2} M$. For a $3 \times 3$ matrix ($n=3$), this simplifies to $\operatorname{adj}(\operatorname{adj}(M)) = (\det M) M$.
• Trigonometric Identities:
  • $\sin A + \sin(A+2\pi/3) + \sin(A+4\pi/3) = 0$
  • $\cos A + \cos(A+2\pi/3) + \cos(A+4\pi/3) = 0$
  • $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$
  • $\sin\theta \sin(\theta+2\pi/3) \sin(\theta+4\pi/3) = -\frac{1}{4}\sin(3\theta)$

2. Understanding the Given Information

The matrix $R$ is a diagonal matrix: $R=\left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right)$ It is specified that $R$ is a non-zero matrix. The trigonometric condition is given as: $x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0$ Let's denote this common non-zero value by $k$. So, $k \neq 0$. From this condition, we can express $x, y, z$ as: $x = \frac{k}{\sin \theta}$ $y = \frac{k}{\sin \left(\theta+\frac{2 \pi}{3}\right)}$ $z = \frac{k}{\sin \left(\theta+\frac{4 \pi}{3}\right)}$ Since $k \neq 0$, it implies that $\sin \theta \neq 0$, $\sin(\theta + 2\pi/3) \neq 0$, and $\sin(\theta + 4\pi/3) \neq 0$. This ensures that $x, y, z$ are all well-defined and non-zero. The condition $\theta \in (0, 2\pi)$ combined with these non-zero sine values implies that $\theta$ cannot be $n\pi$, $n\pi - 2\pi/3$, or $n\pi - 4\pi/3$ for any integer $n$. Specifically, $\theta \neq \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$. This further implies that $\sin(3\theta) \neq 0$.

3. Evaluating Statement (I): Trace $(R)=0$

Goal: Determine if $x+y+z=0$. We substitute the expressions for $x, y, z$ into the trace formula: $\text{Trace}(R) = x+y+z = \frac{k}{\sin \theta} + \frac{k}{\sin \left(\theta+\frac{2 \pi}{3}\right)} + \frac{k}{\sin \left(\theta+\frac{4 \pi}{3}\right)}$ Factor out $k$: $\text{Trace}(R) = k \left( \frac{1}{\sin \theta} + \frac{1}{\sin \left(\theta+\frac{2 \pi}{3}\right)} + \frac{1}{\sin \left(\theta+\frac{4 \pi}{3}\right)} \right)$ To combine the terms within the parenthesis, we find a common denominator: $\text{Trace}(R) = k \frac{\sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right) + \sin \theta \sin \left(\theta+\frac{4 \pi}{3}\right) + \sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right)}{\sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right)}$ Let's evaluate the numerator, $N$: $N = \sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right) + \sin \theta \sin \left(\theta+\frac{4 \pi}{3}\right) + \sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right)$ We use the product-to-sum identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$. Multiplying $N$ by 2: $2N = 2\sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right) + 2\sin \theta \sin \left(\theta+\frac{4 \pi}{3}\right) + 2\sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right)$ Applying the identity to each term:

• $2\sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right) = \cos\left(\left(\theta+\frac{2 \pi}{3}\right)-\left(\theta+\frac{4 \pi}{3}\right)\right) - \cos\left(\left(\theta+\frac{2 \pi}{3}\right)+\left(\theta+\frac{4 \pi}{3}\right)\right)$ $= \cos\left(-\frac{2\pi}{3}\right) - \cos(2\theta+2\pi) = -\frac{1}{2} - \cos(2\theta)$ (since $\cos(X+2\pi)=\cos X$)
• $2\sin \theta \sin \left(\theta+\frac{4 \pi}{3}\right) = \cos\left(\theta-\left(\theta+\frac{4 \pi}{3}\right)\right) - \cos\left(\theta+\left(\theta+\frac{4 \pi}{3}\right)\right)$ $= \cos\left(-\frac{4\pi}{3}\right) - \cos\left(2\theta+\frac{4\pi}{3}\right) = -\frac{1}{2} - \cos\left(2\theta+\frac{4\pi}{3}\right)$
• $2\sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right) = \cos\left(\theta-\left(\theta+\frac{2 \pi}{3}\right)\right) - \cos\left(\theta+\left(\theta+\frac{2 \pi}{3}\right)\right)$ $= \cos\left(-\frac{2\pi}{3}\right) - \cos\left(2\theta+\frac{2\pi}{3}\right) = -\frac{1}{2} - \cos\left(2\theta+\frac{2\pi}{3}\right)$

Summing these three results for $2N$: $2N = \left(-\frac{1}{2} - \cos(2\theta)\right) + \left(-\frac{1}{2} - \cos\left(2\theta+\frac{4\pi}{3}\right)\right) + \left(-\frac{1}{2} - \cos\left(2\theta+\frac{2\pi}{3}\right)\right)$ $2N = -\frac{3}{2} - \left( \cos(2\theta) + \cos\left(2\theta+\frac{2\pi}{3}\right) + \cos\left(2\theta+\frac{4\pi}{3}\right) \right)$ Using the trigonometric identity $\cos A + \cos(A+2\pi/3) + \cos(A+4\pi/3) = 0$ (with $A=2\theta$): $2N = -\frac{3}{2} - 0 = -\frac{3}{2}$ So, the numerator $N = -\frac{3}{4}$.

Now, let's evaluate the denominator, $D = \sin \theta \sin \left(\theta+\frac{2 \pi}{3}\right) \sin \left(\theta+\frac{4 \pi}{3}\right)$. This is a standard identity: $D = -\frac{1}{4}\sin(3\theta)$. Therefore, $\text{Trace}(R) = k \frac{-\frac{3}{4}}{-\frac{1}{4}\sin(3\theta)} = k \frac{3}{\sin(3\theta)}$ Since $k \neq 0$ (given) and $\sin(3\theta) \neq 0$ (as established from $x,y,z$ being well-defined), $\text{Trace}(R) = \frac{3k}{\sin(3\theta)}$ cannot be zero. For example, if we take $\theta=\pi/2$, then $x=k$, $y=-2k$, $z=-2k$. Trace$(R) = k-2k-2k = -3k \neq 0$. Thus, Statement (I) is False.

4. Evaluating Statement (II): If trace $(\operatorname{adj}(\operatorname{adj}(R))=0$, then $R$ has exactly one non-zero entry.

Goal: Analyze the condition $\text{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0$ and its implication. For a $3 \times 3$ matrix $R$, we use the property $\operatorname{adj}(\operatorname{adj}(R)) = (\det R) R$. First, calculate the determinant of $R$: $\det(R) = \det \left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right) = xyz$ Now substitute this into the adjugate property: $\operatorname{adj}(\operatorname{adj}(R)) = (xyz) R = (xyz) \left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right) = \left(\begin{array}{ccc}x^2yz & 0 & 0 \\ 0 & xy^2z & 0 \\ 0 & 0 & xyz^2\end{array}\right)$ Next, we find the trace of this matrix: $\text{Trace}(\operatorname{adj}(\operatorname{adj}(R))) = x^2yz + xy^2z + xyz^2$ Factor out $xyz$: $\text{Trace}(\operatorname{adj}(\operatorname{adj}(R))) = xyz(x+y+z)$ The premise of Statement (II) is $\text{Trace}(\operatorname{adj}(\operatorname{adj}(R)))=0$. So, we have $xyz(x+y+z) = 0$. As established in Section 2, the condition $x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0$ implies that $x, y, z$ are all non-zero. Therefore, $xyz \neq 0$. For the product $xyz(x+y+z)$ to be zero, it must be that $x+y+z=0$. So, the premise of statement (II) simplifies to $x+y+z=0$.

Now we evaluate the conclusion of statement (II): "$R$ has exactly one non-zero entry." If $R$ has exactly one non-zero entry, it means two of $x,y,z$ are zero. For example, if $y=0$ and $z=0$, then $R=\text{diag}(x,0,0)$. However, we derived that $x, y, z$ are all non-zero from the given condition. Therefore, $R$ must have three non-zero entries. The conclusion "R has exactly one non-zero entry" contradicts our finding that all entries $x,y,z$ are non-zero. Thus, Statement (II) is False.

5. Conclusion

Based on our derivations:

• Statement (I) is False.
• Statement (II) is False.

Therefore, neither (I) nor (II) is true.

The option that matches this conclusion is (D).

The problem states that the correct answer is (A). This implies that Statement (I) should be true and Statement (II) should be false. Our analysis for Statement (II) being false is consistent with (A). However, our mathematical derivation for Statement (I) strongly indicates it is false, which contradicts the provided correct answer. If Statement (I) were true, it would require $x+y+z=0$, which implies $3k/\sin(3\theta)=0$, forcing $k=0$, which contradicts the problem's condition that $x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0$.

Final Answer based on mathematical derivation: (D) Neither (I) nor (II) is true.