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(D) The given quadratic equation is $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$.Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = - (3 - \l

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$2$

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(D) The given quadratic equation is $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$.Let the roots be $\alpha$ and $\beta$. Then $\alpha + \beta = - (3 - \lambda) = \lambda - 3$ and $\alpha \beta = 2 - \lambda$.The sum of the squares of the roots is $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$.Substituting the values,we get $S = (\lambda - 3)^2 - 2(2 - \lambda)$.$S = \lambda^2 - 6 \lambda + 9 - 4 + 2 \lambda$.$S = \lambda^2 - 4 \lambda + 5$.To find the least value,we complete the square: $S = (\lambda - 2)^2 + 1$.The expression $S$ has the least value when $\lambda - 2 = 0$,which gives $\lambda = 2$.

The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is

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