Here's a clear, educational, and well-structured solution for the given problem.

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1. Key Concepts and Formulas

For a system of linear equations in three variables: $\begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned}$ to have infinitely many solutions, the following conditions must be met according to Cramer's Rule:

• Determinant of the coefficient matrix ($D$) must be zero. $D = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| = 0$
• Determinants $D_x$, $D_y$, and $D_z$ must also be zero. These are formed by replacing the respective coefficient column in $D$ with the constant terms ($d_1, d_2, d_3$). $D_x = \left|\begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array}\right| = 0 \quad D_y = \left|\begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array}\right| = 0 \quad D_z = \left|\begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array}\right| = 0$ These four conditions ($D=D_x=D_y=D_z=0$) ensure that the system is consistent and represents coincident planes or planes intersecting along a line.

2. Step-by-Step Solution

The given system of linear equations is: $\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned}$

Step 1: Set up the Determinants We first write down the four determinants based on the given system. This is crucial for translating the problem into a solvable form.

• $D$ (Determinant of coefficients): $D = \left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|$
• $D_1$ (or $D_x$, replacing $x$-coefficients with constants): $D_1 = \left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|$
• $D_2$ (or $D_y$, replacing $y$-coefficients with constants): $D_2 = \left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|$
• $D_3$ (or $D_z$, replacing $z$-coefficients with constants): $D_3 = \left|\begin{array}{ccc} 1 & -2 & -4 \\ 2 & \alpha & 5 \\ 3 & -1 & 3 \end{array}\right|$

Step 2: Calculate $D_2$ and solve for $\beta$ To efficiently find the values of $\alpha$ and $\beta$, it's strategic to start with determinants that involve only one unknown. $D_2$ depends only on $\beta$. We calculate $D_2$ by expanding along the first row and set it to zero. $\begin{aligned} D_2 &= 1 \cdot \left| \begin{array}{cc} 5 & 3 \\ 3 & \beta \end{array} \right| - (-4) \cdot \left| \begin{array}{cc} 2 & 3 \\ 3 & \beta \end{array} \right| + 1 \cdot \left| \begin{array}{cc} 2 & 5 \\ 3 & 3 \end{array} \right| \\ &= 1(5 \cdot \beta - 3 \cdot 3) + 4(2 \cdot \beta - 3 \cdot 3) + 1(2 \cdot 3 - 5 \cdot 3) \\ &= (5 \beta - 9) + 4(2 \beta - 9) + (6 - 15) \\ &= 5 \beta - 9 + 8 \beta - 36 - 9 \\ &= 13 \beta - 54 \end{aligned}$ Setting $D_2 = 0$: $13 \beta - 54 = 0$ $13 \beta = 54$ $\beta = \frac{54}{13}$ We have found the value of $\beta$.

Step 3: Calculate $D_3$ and solve for $\alpha$ Similarly, $D_3$ depends only on $\alpha$. We calculate $D_3$ by expanding along the first row and set it to zero to find $\alpha$. $\begin{aligned} D_3 &= 1 \cdot \left| \begin{array}{cc} \alpha & 5 \\ -1 & 3 \end{array} \right| - (-2) \cdot \left| \begin{array}{cc} 2 & 5 \\ 3 & 3 \end{array} \right| + (-4) \cdot \left| \begin{array}{cc} 2 & \alpha \\ 3 & -1 \end{array} \right| \\ &= 1(\alpha \cdot 3 - 5 \cdot (-1)) + 2(2 \cdot 3 - 5 \cdot 3) - 4(2 \cdot (-1) - \alpha \cdot 3) \\ &= (3 \alpha + 5) + 2(6 - 15) - 4(-2 - 3 \alpha) \\ &= 3 \alpha + 5 + 2(-9) - 4(-2 - 3 \alpha) \\ &= 3 \alpha + 5 - 18 + 8 + 12 \alpha \\ &= 15 \alpha - 5 \end{aligned}$ Setting $D_3 = 0$: $15 \alpha - 5 = 0$ $15 \alpha = 5$ $\alpha = \frac{5}{15}$ $\alpha = \frac{1}{3}$ We have found the value of $\alpha$.

Step 4: Verify with $D=0$ For infinitely many solutions, $D$ must also be zero. We substitute the values of $\alpha = \frac{1}{3}$ and $\beta = \frac{54}{13}$ into the expression for $D$ and check if it evaluates to zero. $\begin{aligned} D &= \left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ &= 1(\alpha\beta+3) + 2(2\beta-9) + 1(-2-3\alpha) \\ &= \alpha\beta - 3\alpha + 4\beta - 17 \end{aligned}$ Substitute $\alpha = \frac{1}{3}$ and $\beta = \frac{54}{13}$: $\begin{aligned} D &= \left(\frac{1}{3}\right)\left(\frac{54}{13}\right) - 3\left(\frac{1}{3}\right) + 4\left(\frac{54}{13}\right) - 17 \\ &= \frac{18}{13} - 1 + \frac{216}{13} - 17 \\ &= \frac{18 + 216}{13} - (1 + 17) \\ &= \frac{234}{13} - 18 \\ &= 18 - 18 \\ &= 0 \end{aligned}$ Since $D=0$ is satisfied, our values for $\alpha$ and $\beta$ are consistent.

Step 5: Verify with $D_1=0$ (Optional but Recommended) As a final check, we can verify that $D_1$ also evaluates to zero with our values of $\alpha$ and $\beta$. $D_1 = \left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|$ Expand $D_1$ along the first row: $D_1 = -4(\alpha \beta - 3(-1)) - (-2)(5\beta - 3(3)) + 1(5(-1) - \alpha(3))$ $D_1 = -4(\alpha \beta + 3) + 2(5\beta - 9) + (-5 - 3\alpha)$ Substitute $\alpha = \frac{1}{3}$ and $\beta = \frac{54}{13}$. First, $\alpha\beta = \frac{1}{3} \cdot \frac{54}{13} = \frac{18}{13}$. $\begin{aligned} D_1 &= -4\left(\frac{18}{13} + 3\right) + 2\left(5 \cdot \frac{54}{13} - 9\right) + \left(-5 - 3 \cdot \frac{1}{3}\right) \\ &= -4\left(\frac{18+39}{13}\right) + 2\left(\frac{270-117}{13}\right) + (-5 - 1) \\ &= -4\left(\frac{57}{13}\right) + 2\left(\frac{153}{13}\right) - 6 \\ &= -\frac{228}{13} + \frac{306}{13} - \frac{78}{13} \\ &= \frac{306 - 228 - 78}{13} \\ &= \frac{0}{13} \\ &= 0 \end{aligned}$ All conditions are satisfied, confirming our values for $\alpha$ and $\beta$.

Step 6: Calculate the final expression The question asks for the value of $12 \alpha + 13 \beta$. Substitute $\alpha = \frac{1}{3}$ and $\beta = \frac{54}{13}$: $12 \alpha + 13 \beta = 12 \left(\frac{1}{3}\right) + 13 \left(\frac{54}{13}\right)$ $= 4 + 54$ $= 58$

3. Common Mistakes & Tips

• Determinant Calculation Errors: The most frequent mistake is arithmetic or sign errors while expanding $3 \times 3$ determinants. Double-check each expansion carefully.
• Forgetting All Conditions: For infinitely many solutions, $D=0$ is necessary but not sufficient. All $D_x, D_y, D_z$ must also be zero. Failing to use these additional conditions can lead to incorrect values for the parameters.
• Strategic Choice: When multiple determinants need to be zero, prioritize calculating those that involve fewer unknown variables (like $D_2$ and $D_3$ in this problem) to simplify the solving process.
• Verification: If time permits, use the remaining determinant conditions ($D$ or $D_1$) to verify the derived parameter values. This significantly increases confidence in your answer.

4. Summary

To solve a system of linear equations with infinitely many solutions, we apply Cramer's Rule, which states that the determinant of the coefficient matrix ($D$) and all three modified determinants ($D_x, D_y, D_z$) must be zero. By systematically calculating $D_2$ and $D_3$ and setting them to zero, we found $\beta = \frac{54}{13}$ and $\alpha = \frac{1}{3}$. These values were then verified to satisfy $D=0$ and $D_1=0$. Finally, substituting these values into the required expression $12\alpha + 13\beta$ yielded $58$.

5. Final Answer

The value of $12 \alpha + 13 \beta$ is $58$. The final answer is $\boxed{58}$ which corresponds to option (D).