1. Write the system and solve in terms of $\alpha$

The system is

From the third equation, $x=1-2z.$

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

Now substitute $x=1-2z$ and $y=1+5z$ into the first equation: $(1-2z)+(1+5z)+\alpha z=2$ $2+(3+\alpha)z=2$ $(3+\alpha)z=0.$

For a unique solution, we must have $3+\alpha \ne 0,$ so that $z=0.$ Then

Thus the unique solution is $(x^*,y^*,z^*)=(1,1,0), \quad \alpha\ne -3.$

2. Use the collinearity condition

The given points are: $A=(\alpha,x^*)=(\alpha,1),$ $B=(y^*,\alpha)=(1,\alpha),$ $C=(x^*,-y^*)=(1,-1).$

So we need the points $(\alpha,1),\ (1,\alpha),\ (1,-1)$ to be collinear.

Since $B=(1,\alpha)$ and $C=(1,-1)$ have the same $x$-coordinate, the line through $B$ and $C$ is the vertical line $x=1.$ For $A=(\alpha,1)$ to also lie on this line, we need $\alpha=1.$

This satisfies the unique solution condition $\alpha\ne -3$.

So the only possible value is $\alpha=1.$

3. Find the required sum

Sum of absolute values of all possible values of $\alpha$: $|1|=1.$

4. Check options

The correct value is $1.$ So the correct option is:

• D: 1

5. Compare with stored answer

Stored correct answer is C: 2, but our derivation gives D: 1.

Hence, I disagree with the stored answer.