If $$(1, 5, 35), (7, 5, 5), (1, \lambda, 7)$$ and $$(2\lambda, 1, 2)$$ are coplanar, then the sum of all possible values of $$\lambda$$ is:

We are given four coplanar points $$A = (1, 5, 35)$$, $$B = (7, 5, 5)$$, $$C = (1, \lambda, 7)$$, and $$D = (2\lambda, 1, 2)$$.

For four points to be coplanar, the vectors $$\vec{AB}$$, $$\vec{AC}$$, and $$\vec{AD}$$ must have zero scalar triple product. We compute these vectors:

$$\vec{AB} = (7-1, 5-5, 5-35) = (6, 0, -30)$$

$$\vec{AC} = (1-1, \lambda-5, 7-35) = (0, \lambda-5, -28)$$

$$\vec{AD} = (2\lambda-1, 1-5, 2-35) = (2\lambda-1, -4, -33)$$

Setting the scalar triple product to zero:

$$\begin{vmatrix} 6 & 0 & -30 \\ 0 & \lambda-5 & -28 \\ 2\lambda-1 & -4 & -33 \end{vmatrix} = 0$$

Expanding along the first row:

$$6[(\lambda-5)(-33) - (-28)(-4)] - 0 + (-30)[0 \cdot (-4) - (\lambda-5)(2\lambda-1)] = 0$$

$$6[-33\lambda + 165 - 112] - 30[-(\lambda-5)(2\lambda-1)] = 0$$

$$6[-33\lambda + 53] + 30(\lambda-5)(2\lambda-1) = 0$$

$$-198\lambda + 318 + 30(2\lambda^2 - 11\lambda + 5) = 0$$

$$-198\lambda + 318 + 60\lambda^2 - 330\lambda + 150 = 0$$

$$60\lambda^2 - 528\lambda + 468 = 0$$

Dividing by 12: $$5\lambda^2 - 44\lambda + 39 = 0$$.

By Vieta's formulas, the sum of all possible values of $$\lambda$$ is $$\frac{44}{5}$$.

The answer is Option 1: $$\frac{44}{5}$$.