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

Given: $$x^2 + (3-\lambda)x + 2 = \lambda$$

$$\Rightarrow$$ $$x^2 + (3-\lambda)x + 2 - \lambda = 0$$

Let $$\alpha$$ and $$\beta$$ be the roots of the given equation.

$$\Rightarrow$$ $$\alpha + \beta = -(3 - \lambda) = (\lambda - 3)$$

$$\Rightarrow$$ $$\alpha\beta = 2 - \lambda$$

$$\Rightarrow$$ $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

$$\Rightarrow$$ $$\alpha^2 + \beta^2 = (\lambda - 3)^2 - 2(2 - \lambda) = \lambda^2 - 4\lambda + 5 = (\lambda - 2)^2 + 1$$

We have given that the value of $$\alpha^2 + \beta^2$$ is least. It will be least when $$(\lambda - 2)^2 = 0$$

This occurs at $$\lambda = 2$$

Hence, option A is the correct choice.