$$D = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 3 & 2 \cr 2 & 3 & {{\alpha ^2} - 1} \cr } } \right|$$

D = 3$$a$$² $$-$$ 3 $$-$$ 6 $$-$$ 2$$a$$² + 2 + 4 + 2$$a$$² $$-$$ 2 $$-$$ 4

D = ($$a$$² $$-$$ 3)

When D $$ \ne $$ 0 then system of equiation has unique solution.

$$ \therefore $$  3($$a$$² $$-$$ 3) $$ \ne $$ 0

$$ \Rightarrow $$  $$\left| a \right|$$ $$ \ne \sqrt 3 $$

When $$3({a^2} - 3) = 0$$

$$ \Rightarrow $$  $$\left| a \right| = \sqrt 3 $$ then D = 0

If D = 0 then two cases possible

(1)  System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D₁ = $$\left| {\matrix{ 2 & 1 & 1 \cr 5 & 3 & 2 \cr {a + 1} & 3 & {{a^2} - 1} \cr } } \right| = {a^2} - a + 1$$

D₂ = $$\left| {\matrix{ 1 & 2 & 1 \cr 2 & 5 & 2 \cr 2 & {a + 1} & {{a^2} - 1} \cr } } \right| = {a^2} - 3$$

D₃ = $$\left| {\matrix{ 1 & 1 & 2 \cr 2 & 3 & 5 \cr 2 & 3 & {a + 1} \cr } } \right| = a - 4$$

System of equation will have infinite solution if D₁ = D₂ = D₃ = 0.

And system of equation will have no solution if at last one of D₁, D₂, D₂ is non zero.

At $$\left| a \right| = \sqrt 3 $$ we get D = 0

But D₁ = 3 $$ \pm $$ $$\sqrt 3 + 1$$ $$ \ne $$ 0

and D₃ = $$ \pm $$ $$\sqrt 3 - 4$$ $$ \ne $$ 0

So, system of equations has no solution at $$\left| a \right| = \sqrt 3 $$ then system is in consistent