If $$\langle 2, 3, 9 \rangle$$, $$\langle 5, 2, 1 \rangle$$, $$\langle 1, \lambda, 8 \rangle$$ and $$\langle \lambda, 2, 3 \rangle$$ are coplanar, then the product of all possible values of $$\lambda$$ is

We are given four vectors $$\vec{u} = \langle 2, 3, 9 \rangle$$, $$\vec{v} = \langle 5, 2, 1 \rangle$$, $$\vec{w} = \langle 1, \lambda, 8 \rangle$$, and $$\vec{z} = \langle \lambda, 2, 3 \rangle$$ that are coplanar. Since these are position vectors (or free vectors) that are coplanar, one of them can be expressed as a linear combination of the others. Equivalently, since four vectors in $$\mathbb{R}^3$$ are always linearly dependent, we interpret coplanarity as the condition that the vectors $$\vec{v} - \vec{u}$$, $$\vec{w} - \vec{u}$$, and $$\vec{z} - \vec{u}$$ are coplanar, meaning their scalar triple product is zero.

We compute $$\vec{v} - \vec{u} = \langle 3, -1, -8 \rangle$$, $$\vec{w} - \vec{u} = \langle -1, \lambda - 3, -1 \rangle$$, and $$\vec{z} - \vec{u} = \langle \lambda - 2, -1, -6 \rangle$$.

The scalar triple product equals the determinant: $$\begin{vmatrix} 3 & -1 & -8 \\ -1 & \lambda - 3 & -1 \\ \lambda - 2 & -1 & -6 \end{vmatrix} = 0$$.

Expanding along the first row: $$3[(\lambda - 3)(-6) - (-1)(-1)] - (-1)[(-1)(-6) - (-1)(\lambda - 2)] + (-8)[(-1)(-1) - (\lambda - 3)(\lambda - 2)] = 0$$.

We simplify each term. The first term gives $$3[-6\lambda + 18 - 1] = 3(-6\lambda + 17) = -18\lambda + 51$$. The second term gives $$1[6 - (-\lambda + 2)] = 1[6 + \lambda - 2] = \lambda + 4$$. The third term gives $$-8[1 - (\lambda^2 - 5\lambda + 6)] = -8[1 - \lambda^2 + 5\lambda - 6] = -8[-\lambda^2 + 5\lambda - 5] = 8\lambda^2 - 40\lambda + 40$$.

Adding all terms: $$-18\lambda + 51 + \lambda + 4 + 8\lambda^2 - 40\lambda + 40 = 0$$, which gives $$8\lambda^2 - 57\lambda + 95 = 0$$.

Using the quadratic formula: $$\lambda = \frac{57 \pm \sqrt{3249 - 3040}}{16} = \frac{57 \pm \sqrt{209}}{16}$$. However, we need the product of the roots. By Vieta's formulas, the product of all possible values of $$\lambda$$ is $$\frac{95}{8}$$.

Hence, the correct answer is Option D.