Let S be the set of all real values of $$\lambda$$ such that a plane passing through the points $$(-\lambda^2, 1, 1)$$, $$(1, -\lambda^2, 1)$$ and $$(1, 1, -\lambda^2)$$ also passes through the point $$(-1, -1, 1)$$. Then S is equal to:

We have three given points $$P(-\lambda^{2},\,1,\,1),\;Q(1,\,-\lambda^{2},\,1),\;R(1,\,1,\,-\lambda^{2}).$$ Any plane through these three points must also pass through the point $$S(-1,\,-1,\,1).$$ We shall translate this geometric condition into an algebraic equation in $$\lambda$$.

First we determine a normal vector to the required plane. To do that we form two non-parallel direction vectors lying on the plane:

Vector $$\overrightarrow{PQ}=Q-P$$ is $$\overrightarrow{PQ}=(1-(-\lambda^{2}),\;-\lambda^{2}-1,\;1-1) =(1+\lambda^{2},\;-\lambda^{2}-1,\;0).$$

Vector $$\overrightarrow{PR}=R-P$$ is $$\overrightarrow{PR}=(1-(-\lambda^{2}),\;1-1,\;-\lambda^{2}-1) =(1+\lambda^{2},\;0,\;-\lambda^{2}-1).$$

The normal vector $$\mathbf{n}$$ to the plane is given by the cross product $$\mathbf{n}= \overrightarrow{PQ}\times\overrightarrow{PR}.$$

We compute the cross product using the determinant formula $$\overrightarrow{a}\times\overrightarrow{b}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3} \end{vmatrix}.$$ So

$$\mathbf{n}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1+\lambda^{2}&-\lambda^{2}-1&0\\ 1+\lambda^{2}&0&-\lambda^{2}-1 \end{vmatrix} = \mathbf{i}\Bigl((-\lambda^{2}-1)(-\lambda^{2}-1)-0\Bigr) - \mathbf{j}\Bigl((1+\lambda^{2})(-\lambda^{2}-1)-0\Bigr) + \mathbf{k}\Bigl((1+\lambda^{2})\cdot0-(1+\lambda^{2})(-\lambda^{2}-1)\Bigr).$$

Simplifying step by step, notice that $$-\lambda^{2}-1=-(1+\lambda^{2}).$$ Put $$a=1+\lambda^{2}\;,\quad b=-(1+\lambda^{2})=-a.$$ Then

$$\mathbf{n}=(b\cdot b,\;-a\cdot b,\;-a\cdot b) =\bigl((-a)(-a),\;-a(-a),\;-a(-a)\bigr) =(a^{2},\,a^{2},\,a^{2}).$$

Since $$a^{2}\neq 0$$ for every real $$\lambda$$, a convenient normal vector is $$\mathbf{n}=(1,\,1,\,1)$$ (we divide by $$a^{2}$$ to keep coefficients simple).

With normal vector $$(1,\,1,\,1)$$, the equation of the plane through point $$P(-\lambda^{2},1,1)$$ is written using the point-normal form $$(x-x_{0})+(y-y_{0})+(z-z_{0})=0.$$ Substituting $$(x_{0},y_{0},z_{0})=(-\lambda^{2},\,1,\,1)$$, we have

$$\bigl(x+ \lambda^{2}\bigr)+\bigl(y-1\bigr)+\bigl(z-1\bigr)=0.$$ Collecting like terms,

$$x+y+z+\lambda^{2}-2=0$$ or equivalently $$x+y+z=2-\lambda^{2}. \quad -(1)$$

The extra condition is that this same plane must contain the point $$S(-1,-1,1).$$ Substituting $$x=-1,\;y=-1,\;z=1$$ into equation (1):

$$(-1)+(-1)+1=2-\lambda^{2}.$$ The left-hand side simplifies to $$-1.$$ So we obtain

$$-1=2-\lambda^{2}.$$ Rearranging,

$$\lambda^{2}=2+1=3.$$

Taking square roots (and remembering that both positive and negative roots are real possibilities),

$$\lambda=\sqrt{3}\quad\text{or}\quad \lambda=-\sqrt{3}.$$

Thus the required set of real values is $$S=\{\sqrt{3},-\sqrt{3}\}.$$

Hence, the correct answer is Option D.