This problem requires a solid understanding of homogeneous systems of linear equations and their conditions for non-trivial solutions, combined with proficiency in trigonometric identities and solving trigonometric equations.

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1. Fundamental Concept: Condition for Non-Trivial Solutions

A system of linear equations is classified as homogeneous if all the constant terms on the right-hand side are zero. For the given system:

$$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$$

The solution $x=0, y=0, z=0$ is always valid and is known as the trivial solution. However, the problem states that the system has a non-trivial solution, meaning there exists at least one solution where not all $x, y, z$ are zero.

Key Concept: A homogeneous system of linear equations has a non-trivial solution if and only if the determinant of its coefficient matrix is equal to zero.

First, let's form the coefficient matrix $A$ from the given system:

$$A = \begin{pmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{pmatrix}$$

For a non-trivial solution to exist, we must enforce the condition $\det(A) = 0$.

2. Setting Up the Determinant Equation

We equate the determinant of the coefficient matrix to zero:

$$\left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0$$

3. Evaluating the Determinant and Forming a Trigonometric Equation

To calculate the $3 \times 3$ determinant, we'll use the cofactor expansion method along the first column, as it contains '1's, which often simplifies calculations.

$$\det(A) = 1 \cdot \left|\begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right| - 1 \cdot \left|\begin{array}{cc} \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ \sin \alpha & -\cos \alpha \end{array}\right| + 1 \cdot \left|\begin{array}{cc} \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right| = 0$$

Let's evaluate each $2 \times 2$ determinant systematically:

1. First term (coefficient of the first '1'): $1 \cdot [(\cos \alpha)(-\cos \alpha) - (\sin \alpha)(\sin \alpha)]$ $= 1 \cdot [-\cos^2 \alpha - \sin^2 \alpha]$ $= -(\cos^2 \alpha + \sin^2 \alpha)$ Using the fundamental trigonometric identity $\cos^2 \alpha + \sin^2 \alpha = 1$, this term simplifies to $-1$.

2. Second term (coefficient of the second '1', with a negative sign as per cofactor expansion): $-1 \cdot [(\sqrt{2} \sin \alpha)(-\cos \alpha) - (\sqrt{2} \cos \alpha)(\sin \alpha)]$ $= -1 \cdot [-\sqrt{2} \sin \alpha \cos \alpha - \sqrt{2} \sin \alpha \cos \alpha]$ $= -1 \cdot [-2\sqrt{2} \sin \alpha \cos \alpha]$ $= 2\sqrt{2} \sin \alpha \cos \alpha$

3. Third term (coefficient of the third '1'): $+1 \cdot [(\sqrt{2} \sin \alpha)(\sin \alpha) - (\sqrt{2} \cos \alpha)(\cos \alpha)]$ $= 1 \cdot [\sqrt{2} \sin^2 \alpha - \sqrt{2} \cos^2 \alpha]$ $= \sqrt{2}(\sin^2 \alpha - \cos^2 \alpha)$

Now, substitute these simplified terms back into the determinant equation: $-1 + 2\sqrt{2} \sin \alpha \cos \alpha + \sqrt{2}(\sin^2 \alpha - \cos^2 \alpha) = 0$

To simplify this trigonometric expression further, we employ double angle identities:

• $\sin 2\alpha = 2 \sin \alpha \cos \alpha$
• $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$ (which implies $\sin^2 \alpha - \cos^2 \alpha = -\cos 2\alpha$)

Substituting these identities into the equation: $-1 + \sqrt{2}(2 \sin \alpha \cos \alpha) + \sqrt{2}(-\cos 2\alpha) = 0$ $-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0$ Rearranging the terms, we obtain a standard form trigonometric equation: $\sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 1$

Tip: Be very careful with signs, especially when expanding determinants and applying trigonometric identities. A common mistake is misplacing a negative sign.

4. Solving the Trigonometric Equation

We need to solve the equation $\sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 1$. This is an equation of the form $a \sin x + b \cos x = c$. To solve such equations, we transform the left-hand side into a single trigonometric function using the compound angle formula, typically $R \sin(x - \phi)$ or $R \cos(x + \phi)$. Here, $a = \sqrt{2}$ and $b = -\sqrt{2}$. We calculate $R = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2+2} = \sqrt{4} = 2$.

Divide the entire equation by $R=2$: $\frac{\sqrt{2}}{2} \sin 2\alpha - \frac{\sqrt{2}}{2} \cos 2\alpha = \frac{1}{2}$ $\frac{1}{\sqrt{2}} \sin 2\alpha - \frac{1}{\sqrt{2}} \cos 2\alpha = \frac{1}{2}$ Recognize that $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. Substitute these values: $\sin 2\alpha \cos \frac{\pi}{4} - \cos 2\alpha \sin \frac{\pi}{4} = \frac{1}{2}$ This expression perfectly matches the compound angle formula for $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Here, $A = 2\alpha$ and $B = \frac{\pi}{4}$. So, the equation simplifies to: $\sin \left(2\alpha - \frac{\pi}{4}\right) = \frac{1}{2}$ We know that $\sin \frac{\pi}{6} = \frac{1}{2}$. Thus, we can write: $\sin \left(2\alpha - \frac{\pi}{4}\right) = \sin \frac{\pi}{6}$ The general solution for $\sin x = \sin y$ is given by $x = n\pi + (-1)^n y$, where $n$ is an integer. Applying this general solution to our equation: $2\alpha - \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z}$

5. Finding the Specific Value of $\alpha$ in the Given Interval

The problem specifies that $\alpha \in \left(0, \frac{\pi}{2}\right)$. We need to find the value of $\alpha$ that satisfies this condition from the general solution.

Let's test different integer values for $n$:

• Case 1: $n=0$ $2\alpha - \frac{\pi}{4} = 0 \cdot \pi + (-1)^0 \frac{\pi}{6}$ $2\alpha - \frac{\pi}{4} = \frac{\pi}{6}$ Add $\frac{\pi}{4}$ to both sides: $2\alpha = \frac{\pi}{6} + \frac{\pi}{4}$ Find a common denominator (12): $2\alpha = \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{5\pi}{12}$ Divide by 2: $\alpha = \frac{5\pi}{24}$ Let's check if this value lies in the interval $\left(0, \frac{\pi}{2}\right)$. $\frac{5\pi}{24} > 0$ is true. For the upper bound, $\frac{5\pi}{24} < \frac{\pi}{2}$ implies $\frac{5}{24} < \frac{1}{2}$, which means $5 < 12$. This is true. So, $\alpha = \frac{5\pi}{24}$ is a valid solution within the given interval.

• Case 2: $n=1$ $2\alpha - \frac{\pi}{4} = 1 \cdot \pi + (-1)^1 \frac{\pi}{6}$ $2\alpha - \frac{\pi}{4} = \pi - \frac{\pi}{6}$ $2\alpha - \frac{\pi}{4} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$ Add $\frac{\pi}{4}$ to both sides: $2\alpha = \frac{5\pi}{6} + \frac{\pi}{4}$ Find a common denominator (12): $2\alpha = \frac{10\pi}{12} + \frac{3\pi}{12} = \frac{13\pi}{12}$ Divide by 2: $\alpha = \frac{13\pi}{24}$ Let's check if this value lies in the interval $\left(0, \frac{\pi}{2}\right)$. $\frac{13\pi}{24} > 0$ is true. For the upper bound, $\frac{13\pi}{24} < \frac{\pi}{2}$ implies $\frac{13}{24} < \frac{1}{2}$, which means $13 < 12$. This is false. So, $\alpha = \frac{13\pi}{24}$ is NOT a valid solution in the given interval.

• Case 3: $n=-1$ $2\alpha - \frac{\pi}{4} = (-1) \cdot \pi + (-1)^{-1} \frac{\pi}{6}$ $2\alpha - \frac{\pi}{4} = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6}$ $2\alpha = \frac{\pi}{4} - \frac{7\pi}{6} = \frac{3\pi - 14\pi}{12} = -\frac{11\pi}{12}$ $\alpha = -\frac{11\pi}{24}$ This value is negative and thus not in the interval $\left(0, \frac{\pi}{2}\right)$.

Therefore, the only value of $\alpha$ in the interval $\left(0, \frac{\pi}{2}\right)$ that satisfies the condition is $\alpha = \frac{5\pi}{24}$.

The correct option is (A).

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Summary and Key Takeaways

1. Non-Trivial Solutions for Homogeneous Systems: Remember the critical condition: $\det(A) = 0$ for a homogeneous system to have non-trivial solutions. This is a fundamental concept in linear algebra.
2. Determinant Calculation: Practice calculating $3 \times 3$ determinants efficiently and accurately. Row/column operations can sometimes simplify the calculation, but direct cofactor expansion is also a reliable method.
3. Trigonometric Identities: Mastery of basic and double-angle trigonometric identities (like $\sin^2\alpha + \cos^2\alpha = 1$, $\sin 2\alpha$, $\cos 2\alpha$) is crucial for simplifying expressions.
4. Solving $a \sin x + b \cos x = c$: This standard form trigonometric equation is best solved by converting the LHS into a single trigonometric function using $R \sin(x \pm \phi)$ or $R \cos(x \pm \phi)$.
5. General Solutions and Interval Checking: Always use the general solution formula for trigonometric equations and then carefully check which values fall within the specified domain or interval given in the problem. This step is vital to avoid selecting incorrect answers.