Let the lines $$\frac{x-1}{\lambda} = \frac{y-2}{1} = \frac{z-3}{2}$$ and $$\frac{x+26}{-2} = \frac{y+18}{3} = \frac{z+28}{\lambda}$$ be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lie on P?

We have lines $$L_1: \frac{x-1}{\lambda} = \frac{y-2}{1} = \frac{z-3}{2}$$ and $$L_2: \frac{x+26}{-2} = \frac{y+18}{3} = \frac{z+28}{\lambda}$$.

For the lines to be coplanar, the determinant formed by the direction vectors and the vector joining the points on the two lines must be zero.

The point on $$L_1$$ is $$(1, 2, 3)$$ and on $$L_2$$ is $$(-26, -18, -28)$$. The vector joining them is $$(-27, -20, -31)$$.

The direction vectors are $$(\lambda, 1, 2)$$ and $$(-2, 3, \lambda)$$.

The coplanarity condition is:

$$\begin{vmatrix} -27 & -20 & -31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda \end{vmatrix} = 0$$

Expanding along the first row:

$$-27(\lambda - 6) + 20(\lambda^2 + 4) - 31(3\lambda + 2) = 0$$

$$-27\lambda + 162 + 20\lambda^2 + 80 - 93\lambda - 62 = 0$$

$$20\lambda^2 - 120\lambda + 180 = 0$$

$$\lambda^2 - 6\lambda + 9 = 0$$

$$(\lambda - 3)^2 = 0$$

So $$\lambda = 3$$.

Now we find the equation of the plane $$P$$ containing both lines. With $$\lambda = 3$$, the direction vectors are $$(3, 1, 2)$$ and $$(-2, 3, 3)$$.

The normal to the plane is their cross product:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ -2 & 3 & 3 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(9+4) + \hat{k}(9+2) = (-3, -13, 11)$$

The plane passes through $$(1, 2, 3)$$:

$$-3(x-1) - 13(y-2) + 11(z-3) = 0$$

$$-3x + 3 - 13y + 26 + 11z - 33 = 0$$

$$-3x - 13y + 11z - 4 = 0$$

Or equivalently: $$3x + 13y - 11z + 4 = 0$$.

Now we check each point:

$$(0, -2, -2)$$: $$0 - 26 + 22 + 4 = 0$$ — lies on $$P$$.

$$(-5, 0, -1)$$: $$-15 + 0 + 11 + 4 = 0$$ — lies on $$P$$.

$$(3, -1, 0)$$: $$9 - 13 + 0 + 4 = 0$$ — lies on $$P$$.

$$(0, 4, 5)$$: $$0 + 52 - 55 + 4 = 1 \neq 0$$ — does NOT lie on $$P$$.

Hence, the correct answer is Option D.