Given quadratic equation,

$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$

Let $$\alpha $$ and $$\beta $$ are the roots of the equation.

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$

and $$\alpha $$$$\beta $$ = $$ - a - 1$$

Now $${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$$

$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$$

$$ \Rightarrow $$ The value of $${\alpha ^2} + {\beta ^2}$$ will be minimum, when $${a - 1}$$ = 0

$$ \Rightarrow $$ $${a = 1}$$