Let the system of linear equations
$$x + y + az = 2$$
$$3x + y + z = 4$$
$$x + 2z = 1$$
have a unique solution $$(x^*, y^*, z^*)$$. If $$((a, x^*), (y^*, \alpha)$$ and $$(x^*, -y^*)$$ are collinear points, then the sum of absolute values of all possible values of $$\alpha$$ is:

Given,

$$x+y+\alpha z=2$$

$$3x+y+z=4$$

$$x+2z=1$$

The coefficient matrix is

$$A=\begin{pmatrix}1&1&\alpha\\3&1&1\\1&0&2\end{pmatrix}$$

For a unique solution,

$$|A|\neq0$$

Now,

$$|A|=\begin{vmatrix}1&1&\alpha\\3&1&1\\1&0&2\end{vmatrix}=-(\alpha+3)$$

Hence,

$$\alpha\neq-3$$

Using Cramer’s rule,

$$\Delta_1=\begin{vmatrix}2&1&\alpha\\4&1&1\\1&0&2\end{vmatrix}=-(\alpha+3)$$

$$\Delta_2=\begin{vmatrix}1&2&\alpha\\3&4&1\\1&1&2\end{vmatrix}=-(\alpha+3)$$

$$\Delta_3=\begin{vmatrix}1&1&2\\3&1&4\\1&0&1\end{vmatrix}=0$$

Therefore,

$$x^*=\frac{\Delta_1}{\Delta}=1,\qquad y^*=\frac{\Delta_2}{\Delta}=1,\qquad z^*=\frac{\Delta_3}{\Delta}=0$$

The given points are

$$ (\alpha,x^*)=(\alpha,1),\qquad (y^*,\alpha)=(1,\alpha),\qquad (x^*,-y^*)=(1,-1) $$

Since they are collinear,

$$\begin{vmatrix}\alpha&1&1\\1&\alpha&1\\1&-1&1\end{vmatrix}=0$$

Expanding,

$$\alpha(\alpha+1)-1(1-1)+1(-1-\alpha)=0$$

$$\alpha^2+\alpha-1-\alpha=0$$

$$\alpha^2-1=0$$

$$(\alpha-1)(\alpha+1)=0$$

Hence,

$$\alpha=\pm1$$

Therefore, the sum of absolute values of all possible values of $$\alpha$$ is

$$|1|+|-1|=2$$

Hence, $$\boxed{2}$$.