Question:

<p><strong>Step 1:</strong> Form the coefficient matrix $A$:</p><p>$$A = \begin{bmatrix} 1+\alpha & \beta & 1 \\ \alpha & 1+\beta & 1 \\ \alpha & \beta & 2 \end{bmatrix}$$</p><p><strong>Step 2:</strong> Calculate $\det(A)$ by performing row operations. $R_2 \to R_2 - R_3$ and $R_1 \to R_1 - R_3$:</p><p>$R_1: (1+\alpha-\alpha)x + (\beta-\beta)y + (1-2)z = (2-2) \Rightarrow x - z = 0$</p><p>$R_2: (\alpha-\alpha)x + (1+\beta-\beta)y + (1-2)z = (3-2) \Rightarrow y - z = 1$</p><p>$R_3: \alpha x + \beta y + 2z = 2$</p><p><strong>Step 3:</strong> Calculate $\det(A)$:</p><p>$$\det(A) = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \alpha & \beta & 2 \end{vmatrix} = 1(2+\beta) - 0 + (-1)(0-\alpha) = 2+\beta+\alpha$$</p><p><strong>Step 4:</strong> For unique solution: $\det(A) \neq 0$</p><p>$$\Rightarrow \alpha + \beta + 2 \neq 0$$</p><p>$$\Rightarrow \alpha + \beta \neq -2$$</p><p><strong>Step 5:</strong> The system has a unique solution for all $(\alpha, \beta)$ except those on the line $\alpha + \beta = -2$. Since this represents infinitely many excluded pairs on a line in $\mathbb{R}^2$, the answer is infinitely many ordered pairs (excluding the line $\alpha + \beta = -2$).</p><p>$\therefore$ Answer: C (Infinitely many)</p>