It is given that

$$\left[ {\matrix{ 1 & \alpha & {{\alpha ^2}} \cr \alpha & 1 & \alpha \cr {{\alpha ^2}} & \alpha & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$$

$$\left[ {\matrix{ 1 & \alpha & {{\alpha ^2}} \cr \alpha & 1 & \alpha \cr {{\alpha ^2}} & \alpha & 1 \cr } } \right] = 0$$

$$ \Rightarrow 1(1 - {\alpha ^2}) - \alpha (\alpha - {\alpha ^3}) + {\alpha ^2}({\alpha ^2} - {\alpha ^2}) = 0$$

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

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

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

$$ \Rightarrow {\alpha ^2} = 1 \Rightarrow \alpha = \pm 1$$

For $$\alpha$$ = 1, the given system of linear equations has no solution.

$$\left[ {\matrix{ { + 1} & { + 1} & { + 1} \cr { + 1} & { + 1} & { + 1} \cr { + 1} & { + 1} & { + 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$$

$$x + y + z = 1$$

$$x + y + z = - 1$$

$$x + y + z = 1$$

Since two planes are parallel. So, $$\alpha$$ = 1 is rejected for $$\alpha$$ = $$-$$1 the given system of linear equations has coincident planes

$$\left[ {\matrix{ 1 & { - 1} & 1 \cr { - 1} & 1 & { - 1} \cr 1 & { - 1} & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$$

$$x - y + z = 1$$

$$ \Rightarrow - x + y - z = - 1 \Rightarrow x - y + z = 1$$

$$x - y + 1 = 1$$

Therefore, $$\alpha$$ = $$-$$1 is accepted. That is,

$$1 + \alpha + {\alpha ^2} = 1 + ( - 1) + {( - 1)^2} = 1 - 1 + 1 = 1$$

$$ \Rightarrow 1 + \alpha + {\alpha ^2} = 1$$