This problem combines concepts from systems of linear equations and coordinate geometry. We need to first determine the conditions for a unique solution of the given system, then find that unique solution $(x^*, y^*, z^*)$, and finally use the collinearity condition for the three given points to find the possible values of $\alpha$.

1. Key Concepts and Formulas

   • Unique Solution for a System of Linear Equations: For a system of linear equations represented in matrix form $AX = B$, a unique solution exists if and only if the determinant of the coefficient matrix $A$ is non-zero, i.e., $\det(A) \neq 0$. If $\det(A) = 0$, the system either has no solution or infinitely many solutions.
   • Collinearity of Three Points: Three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear (lie on the same straight line) if and only if the area of the triangle formed by these points is zero. This condition can be expressed using a determinant: $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$

2. Step-by-Step Solution

   Step 1: Determine the condition for a unique solution. The given system of linear equations is:

   1. $x + y + \alpha z = 2$
   2. $3x + y + z = 4$
   3. $x + 2z = 1$

   First, we write the coefficient matrix $A$: $A = \begin{pmatrix} 1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix}$ Next, we calculate the determinant of $A$ to find the condition for a unique solution. We expand along the first row: $\det(A) = 1 \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} - 1 \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} + \alpha \begin{vmatrix} 3 & 1 \\ 1 & 0 \end{vmatrix}$ $\det(A) = 1(1 \cdot 2 - 0 \cdot 1) - 1(3 \cdot 2 - 1 \cdot 1) + \alpha(3 \cdot 0 - 1 \cdot 1)$ $\det(A) = 1(2) - 1(6 - 1) + \alpha(-1)$ $\det(A) = 2 - 5 - \alpha = -3 - \alpha$ For the system to have a unique solution, $\det(A) \neq 0$. So, $-3 - \alpha \neq 0 \Rightarrow \alpha \neq -3$. This is the condition on $\alpha$.

   Step 2: Determine the unique solution $(x^*, y^*, z^*)$. We use substitution to solve the system for $(x^*, y^*, z^*)$ for $\alpha \neq -3$. From equation (3): $x = 1 - 2z$. (Equation 4)

   Substitute (4) into equation (2): $3(1 - 2z) + y + z = 4$ $3 - 6z + y + z = 4$ $y - 5z = 1$ $y = 1 + 5z$ (Equation 5)

   Substitute (4) and (5) into equation (1): $(1 - 2z) + (1 + 5z) + \alpha z = 2$ $2 + 3z + \alpha z = 2$ $(3 + \alpha)z = 0$

   Since we established that for a unique solution, $\alpha \neq -3$, it means $3 + \alpha \neq 0$. Therefore, for the equation $(3 + \alpha)z = 0$ to hold, we must have $z^* = 0$.

   Now, substitute $z^* = 0$ back into (4) and (5) to find $x^*$ and $y^*$: From (4): $x^* = 1 - 2(0) = 1$. From (5): $y^* = 1 + 5(0) = 1$. Thus, for any $\alpha \neq -3$, the unique solution to the system is $(x^*, y^*, z^*) = (1, 1, 0)$.

   Step 3: Apply the collinearity condition to find $\alpha$. We are given three points: $(\alpha, x^*)$, $(y^*, \alpha)$, and $(x^*, -y^*)$. Substitute the values $x^* = 1$ and $y^* = 1$ into these points: Point 1: $(\alpha, 1)$ Point 2: $(1, \alpha)$ Point 3: $(1, -1)$

   For these three points to be collinear, the determinant condition must be satisfied: $\begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1 \end{vmatrix} = 0$ Now, we evaluate this determinant by expanding along the first row: $\alpha(\alpha \cdot 1 - 1 \cdot (-1)) - 1(1 \cdot 1 - 1 \cdot 1) + 1(1 \cdot (-1) - \alpha \cdot 1) = 0$ $\alpha(\alpha + 1) - 1(1 - 1) + 1(-1 - \alpha) = 0$ $\alpha(\alpha + 1) - 1(0) + (-1 - \alpha) = 0$ $\alpha^2 + \alpha - 4 - \alpha = 0$ $\alpha^2 - 4 = 0$ $(\alpha - 2)(\alpha + 2) = 0$ This gives us two possible values for $\alpha$: $\alpha = 2$ or $\alpha = -2$.

   Step 4: Validate $\alpha$ values and calculate the sum of absolute values. We found the possible values for $\alpha$ are $2$ and $-2$. We also established the condition for a unique solution: $\alpha \neq -3$. Both $\alpha = 2$ and $\alpha = -2$ satisfy the condition $\alpha \neq -3$. Therefore, both values are valid.

   The problem asks for the sum of the absolute values of all possible values of $\alpha$. Sum of absolute values $= |\alpha_1| + |\alpha_2|$ $= |2| + |-2|$ $= 2 + 2 = 4$.

3. Common Mistakes & Tips

   • Always check the condition for a unique solution (i.e., $\det(A) \neq 0$) against the values of $\alpha$ obtained from the collinearity condition. Any $\alpha$ that makes $\det(A) = 0$ must be discarded.
   • Be careful with algebraic manipulations when expanding determinants, as a small sign error or calculation mistake can lead to incorrect values for $\alpha$.
   • Verify your solution for $(x^*, y^*, z^*)$ by plugging the values back into the original system of equations.

4. Summary

   To solve this problem, we first determined the condition for a unique solution of the linear system by ensuring the determinant of the coefficient matrix was non-zero. This yielded $\alpha \neq -3$. Next, we solved the system to find the unique solution $(x^*, y^*, z^*) = (1, 1, 0)$. Finally, we substituted these values into the given points and applied the collinearity condition using a determinant. This led to the equation $\alpha^2 - 4 = 0$, giving $\alpha = 2$ and $\alpha = -2$. Both values satisfy the unique solution condition. The sum of their absolute values is $2 + 2 = 4$.

5. Final Answer

   The final answer is $\boxed{4}$, which corresponds to option (A).