This problem requires a thorough understanding of how to determine the nature of solutions (unique, no solution, or infinitely many solutions) for a system of linear equations. We will use the determinant method, often associated with Cramer's Rule, to analyze the given system under different conditions for $\alpha$ and $\beta$.

1. Key Concepts and Formulas

• System of Linear Equations: A system of $n$ linear equations in $n$ variables can be written in matrix form as $AX=B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constant matrix.
• Determinants:
  • $\Delta = \det(A)$: Determinant of the coefficient matrix.
  • $\Delta_x, \Delta_y, \Delta_z$: Determinants formed by replacing the respective variable's coefficient column in $A$ with the constant terms from $B$.
• Conditions for Solutions:
  • Unique Solution: If $\Delta \neq 0$. (By Cramer's Rule, $x = \frac{\Delta_x}{\Delta}$, $y = \frac{\Delta_y}{\Delta}$, $z = \frac{\Delta_z}{\Delta}$).
  • No Solution: If $\Delta = 0$ AND at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero.
  • Infinitely Many Solutions: If $\Delta = 0$ AND $\Delta_x = \Delta_y = \Delta_z = 0$.

2. Step-by-Step Solution

The given system of linear equations is: $\begin{aligned} \alpha x+y+z&=1 \\ x+\alpha y+z&=1 \\ x+y+\alpha z&=\beta \end{aligned}$

Step 1: Calculate the Determinant of the Coefficient Matrix ($\Delta$)

First, we write the coefficient matrix $A$ and calculate its determinant $\Delta$: $A = \left[\begin{array}{ccc} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right]$ $\Delta = \left|\begin{array}{ccc} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|$ To evaluate $\Delta$, we can apply column operations to simplify it before expansion. Add $C_2$ and $C_3$ to $C_1$: $C_1 \to C_1+C_2+C_3$. $\Delta = \left|\begin{array}{ccc} \alpha+2 & 1 & 1 \\ \alpha+2 & \alpha & 1 \\ \alpha+2 & 1 & \alpha \end{array}\right|$ Factor out $(\alpha+2)$ from $C_1$: $\Delta = (\alpha+2) \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|$ Now, perform row operations: $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$: $\Delta = (\alpha+2) \left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \alpha-1 & 0 \\ 0 & 0 & \alpha-1 \end{array}\right|$ This is an upper triangular matrix, so its determinant is the product of its diagonal elements: $\Delta = (\alpha+2) \cdot 1 \cdot (\alpha-1) \cdot (\alpha-1) = (\alpha+2)(\alpha-1)^2$ The system will not have a unique solution if $\Delta = 0$, which occurs when $(\alpha+2)(\alpha-1)^2 = 0$. This gives $\alpha = 1$ or $\alpha = -2$. We will analyze the options using these critical values.

Step 2: Analyze Option (A)

Statement (A): It has infinitely many solutions if $\alpha=1$ and $\beta=1$.

• Check $\Delta$: For $\alpha=1$, $\Delta = (1+2)(1-1)^2 = 3 \cdot 0^2 = 0$. This condition is necessary for infinitely many solutions.
• Substitute $\alpha=1, \beta=1$ into the system: $\begin{aligned} 1x+y+z&=1 \\ x+1y+z&=1 \\ x+y+1z&=1 \end{aligned}$ All three equations become identical: $x+y+z=1$. This represents a single plane in 3D space. A plane contains infinitely many points, so the system has infinitely many solutions.
• Check $\Delta_x, \Delta_y, \Delta_z$ (optional but good for completeness): For $\alpha=1, \beta=1$: $\Delta_x = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right| = 0 \quad (\text{Columns are identical})$ Similarly, $\Delta_y=0$ and $\Delta_z=0$.
• Conclusion: Since $\Delta=0$ and $\Delta_x=\Delta_y=\Delta_z=0$, the system has infinitely many solutions. Therefore, statement (A) is correct.

Step 3: Analyze Option (B)

Statement (B): It has infinitely many solutions if $\alpha=2$ and $\beta=-1$.

• Check $\Delta$: For $\alpha=2$, $\Delta = (2+2)(2-1)^2 = 4 \cdot 1^2 = 4$.
• Conclusion: Since $\Delta = 4 \neq 0$, the system has a unique solution (by Cramer's Rule), regardless of the value of $\beta$. It cannot have infinitely many solutions. Therefore, statement (B) is NOT correct.

Step 4: Analyze Option (C)

Statement (C): $x+y+z=\frac{3}{4}$ if $\alpha=2$ and $\beta=1$.

• Check $\Delta$: For $\alpha=2$, $\Delta = 4 \neq 0$. This ensures a unique solution exists.
• Substitute $\alpha=2, \beta=1$ into the system: $\begin{aligned} 2x+y+z&=1 \quad &(1) \\ x+2y+z&=1 \quad &(2) \\ x+y+2z&=1 \quad &(3) \end{aligned}$
• Solve for $x+y+z$: Add equations (1), (2), and (3): $(2x+y+z) + (x+2y+z) + (x+y+2z) = 1+1+1$ $(2x+x+x) + (y+2y+y) + (z+z+2z) = 3$ $4x + 4y + 4z = 3$ $4(x+y+z) = 3$ $x+y+z = \frac{3}{4}$
• Conclusion: The sum $x+y+z$ is indeed $\frac{3}{4}$. Therefore, statement (C) is correct.

Step 5: Analyze Option (D)

Statement (D): It has no solution if $\alpha=-2$ and $\beta=1$.

• Check $\Delta$: For $\alpha=-2$, $\Delta = (-2+2)(-2-1)^2 = 0 \cdot (-3)^2 = 0$. This condition is necessary for no solution.
• Substitute $\alpha=-2, \beta=1$ into the system: $\begin{aligned} -2x+y+z&=1 \\ x-2y+z&=1 \\ x+y-2z&=1 \end{aligned}$
• Check $\Delta_x$: We need to calculate at least one of $\Delta_x, \Delta_y, \Delta_z$. Let's calculate $\Delta_x$: $\Delta_x = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{array}\right|$ Expand along the first row: $\begin{aligned} \Delta_x &= 1((-2)(-2) - 1 \cdot 1) - 1(1 \cdot (-2) - 1 \cdot 1) + 1(1 \cdot 1 - (-2) \cdot 1) \\ &= 1(4 - 1) - 1(-2 - 1) + 1(1 + 2) \\ &= 1(3) - 1(-3) + 1(3) \\ &= 3 + 3 + 3 = 9 \end{aligned}$
• Conclusion: Since $\Delta = 0$ and $\Delta_x = 9 \neq 0$, the system has no solution. Therefore, statement (D) is correct.

3. Common Mistakes & Tips

• Determinant Calculation: Be careful with signs and arithmetic when calculating determinants. Using row/column operations can simplify the process and reduce errors.
• Confusing Conditions: Clearly distinguish between the conditions for no solution ($\Delta=0$, at least one $\Delta_i \neq 0$) and infinitely many solutions ($\Delta=0$, and all $\Delta_x, \Delta_y, \Delta_z = 0$).
• Symmetric Systems: For symmetric systems (like in Option C), adding all equations is often a quick way to find $x+y+z$.
• System Consistency: Remember that if $\Delta \neq 0$, the system is always consistent and has a unique solution. Only when $\Delta = 0$ do we need to check $\Delta_x, \Delta_y, \Delta_z$.

4. Summary

We systematically analyzed each option using the determinant method. We first calculated the determinant of the coefficient matrix, $\Delta = (\alpha+2)(\alpha-1)^2$, to identify the critical values of $\alpha$. Then, we evaluated each statement:

• Statement (A) was found to be correct, as for $\alpha=1, \beta=1$, the system reduces to $x+y+z=1$, which has infinitely many solutions.
• Statement (B) was found to be incorrect, as for $\alpha=2$, $\Delta \neq 0$, implying a unique solution, not infinitely many.
• Statement (C) was found to be correct, as for $\alpha=2, \beta=1$, summing the equations directly yields $x+y+z=3/4$.
• Statement (D) was found to be correct, as for $\alpha=-2, \beta=1$, $\Delta=0$ and $\Delta_x \neq 0$, indicating no solution.

The question asks for the statement that is NOT correct. Based on our analysis, statement (B) is the only incorrect statement.

5. Final Answer

The final answer is $\boxed{A}$.