Given system of equations

$$x + y + az = 2$$ ..... (i)

$$3x + y + z = 4$$ ..... (ii)

$$x + 2z = 1$$ ..... (iii)

Solving (i), (ii) and (iii), we get

x = 1, y = 1, z = 0 (and for unique solution a $$\ne$$ $$-$$3)

Now, ($$\alpha$$, 1), (1, $$\alpha$$) and (1, $$-$$1) are collinear

$$\therefore$$ $$\left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & { - 1} & 1 \cr } } \right| = 0$$

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

$$ \Rightarrow {\alpha ^2} - 1 = 0$$

$$\therefore$$ $$\alpha = \, \pm \,1$$

$$\therefore$$ Sum of absolute values of $$\alpha = 1 + 1 = 2$$