1. Given quadratic

We have $3x^2 + (\alpha-6)x + (\alpha+3)=0.$

Let its real roots be $r_1,r_2$.

We need the integral values of $\alpha$ for which the sum of squares of the two real roots is minimum.

2. Condition for real roots

For the roots to be real, discriminant must be non-negative: $\Delta=(\alpha-6)^2-4\cdot 3\cdot (\alpha+3).$

So, $\Delta=\alpha^2-12\alpha+36-12\alpha-36=\alpha^2-24\alpha=\alpha(\alpha-24).$

Hence, $\Delta\ge 0 \iff \alpha(\alpha-24)\ge 0.$

Thus, $\alpha\le 0 \quad \text{or} \quad \alpha\ge 24.$

3. Sum of squares of roots

Using Vieta's formulas: $r_1+r_2=-\frac{\alpha-6}{3}=\frac{6-\alpha}{3},$ $r_1r_2=\frac{\alpha+3}{3}.$

Now, $r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2.$

So, $r_1^2+r_2^2=\left(\frac{6-\alpha}{3}\right)^2-2\left(\frac{\alpha+3}{3}\right).$

Simplify: $r_1^2+r_2^2=\frac{(\alpha-6)^2}{9}-\frac{2(\alpha+3)}{3}.$

Taking LCM $9$, r_1^2+r_2^2=\frac{\alpha^2-12\alpha+36-6\alpha-18}{9}=rac{\alpha^2-18\alpha+18}{9}.

Let $f(\alpha)=\frac{\alpha^2-18\alpha+18}{9}.$

We must minimize $f(\alpha)$ over integral $\alpha$ satisfying $\alpha\le 0 \text{ or } \alpha\ge 24.$

4. Analyze the quadratic

Since $f(\alpha)$ is a quadratic opening upward, its vertex is at $\alpha=\frac{18}{2}=9.$

So the unrestricted minimum occurs at $\alpha=9$, but $9$ does not satisfy the real-root condition.

Thus we minimize over the allowed region: $(-\infty,0] \cup [24,\infty).$

Because the vertex is at $9$, the minimum on the allowed set will occur at the nearest boundary point(s):

• on $\alpha\le 0$, minimum at $\alpha=0$;
• on $\alpha\ge 24$, minimum at $\alpha=24$.

Now compute: $f(0)=\frac{18}{9}=2,$ f(24)=\frac{24^2-18\cdot 24+18}{9}=rac{576-432+18}{9}=\frac{162}{9}=18.

Hence the minimum value is attained at $\alpha=0.$

Since $\alpha=0$ is an integer, the set $S=\{0\}.$

5. Nature of the set $S$

$S$ has exactly one element, so it is a singleton.

Therefore, the correct option is: $\boxed{\text{B}}$

6. Compare with stored correct answer

Stored correct answer: A (empty set)

But our derivation shows $S=\{0\},$ so $S$ is not empty; it is a singleton.

Thus the stored answer appears to be incorrect.