Let the position vectors of the points A, B, C and D be $$5\hat{i} + 5\hat{j} + 2\lambda\hat{k}$$, $$\hat{i} + 2\hat{j} + 3\hat{k}$$, $$-2\hat{i} + \lambda\hat{j} + 4\hat{k}$$ and $$-\hat{i} + 5\hat{j} + 6\hat{k}$$. Let the set $$S = \{\lambda \in \mathbb{R}$$: the points A, B, C and D are coplanar$$\}$$. The $$\sum_{\lambda \in S} (\lambda + 2)^2$$ is equal to

Given position vectors: A = (5,5,2λ), B = (1,2,3), C = (-2,λ,4), D = (-1,5,6).

For coplanarity, the vectors AB, AC, AD must have zero scalar triple product.

$$\vec{AB} = (-4, -3, 3-2\lambda)$$, $$\vec{AC} = (-7, \lambda-5, 4-2\lambda)$$, $$\vec{AD} = (-6, 0, 6-2\lambda)$$.

$$ \begin{vmatrix} -4 & -3 & 3-2\lambda \\ -7 & \lambda-5 & 4-2\lambda \\ -6 & 0 & 6-2\lambda \end{vmatrix} = 0 $$

Expanding along the first row:

$$= -4[(\lambda-5)(6-2\lambda)-0] + 3[(-7)(6-2\lambda)+6(4-2\lambda)] + (3-2\lambda)[0+6(\lambda-5)]$$

$$= -4(\lambda-5)(6-2\lambda) + 3(-42+14\lambda+24-12\lambda) + 6(3-2\lambda)(\lambda-5)$$

$$= -4(\lambda-5)(6-2\lambda) + 3(2\lambda-18) + 6(3-2\lambda)(\lambda-5)$$

$$= (\lambda-5)[-4(6-2\lambda)+6(3-2\lambda)] + 6\lambda-54$$

$$= (\lambda-5)(-24+8\lambda+18-12\lambda) + 6\lambda-54$$

$$= (\lambda-5)(-6-4\lambda) + 6\lambda-54$$

$$= -6\lambda-4\lambda^2+30+20\lambda+6\lambda-54$$

$$= -4\lambda^2+20\lambda-24 = -4(\lambda^2-5\lambda+6) = -4(\lambda-2)(\lambda-3) = 0$$

So $$\lambda = 2$$ or $$\lambda = 3$$.

$$ \sum_{\lambda \in S}(\lambda+2)^2 = (2+2)^2 + (3+2)^2 = 16 + 25 = 41 $$

The correct answer is 41.