The plane containing the line $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$$ and parallel to the line $$\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$$ passes through the point:

The plane contains the line given by $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3}$$. This line passes through the point $$(1, 2, 3)$$ and has a direction vector $$\vec{d_1} = \langle 1, 2, 3 \rangle$$. The plane is also parallel to the line $$\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$$, which has a direction vector $$\vec{d_2} = \langle 1, 1, 4 \rangle$$.

Since the plane contains the first line, it must pass through $$(1, 2, 3)$$ and be parallel to $$\vec{d_1}$$. Additionally, because it is parallel to the second line, it must be parallel to $$\vec{d_2}$$. Therefore, the plane is parallel to both vectors $$\vec{d_1}$$ and $$\vec{d_2}$$. The normal vector $$\vec{n}$$ to the plane is perpendicular to both direction vectors and can be found using their cross product:

$$\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 3 \\ 1 & 1 & 4 \end{vmatrix}$$

Expanding the determinant:

$$\vec{n} = \vec{i}(2 \cdot 4 - 3 \cdot 1) - \vec{j}(1 \cdot 4 - 3 \cdot 1) + \vec{k}(1 \cdot 1 - 2 \cdot 1)$$

Calculating each component:

For the $$\vec{i}$$-component: $$2 \cdot 4 = 8$$, $$3 \cdot 1 = 3$$, so $$8 - 3 = 5$$.

For the $$\vec{j}$$-component: $$1 \cdot 4 = 4$$, $$3 \cdot 1 = 3$$, so $$4 - 3 = 1$$, but since it is subtracted, we have $$-1$$.

For the $$\vec{k}$$-component: $$1 \cdot 1 = 1$$, $$2 \cdot 1 = 2$$, so $$1 - 2 = -1$$.

Thus, $$\vec{n} = \langle 5, -1, -1 \rangle$$.

The plane passes through the point $$(1, 2, 3)$$, so its equation is:

$$\vec{n} \cdot (\langle x, y, z \rangle - \langle 1, 2, 3 \rangle) = 0$$

Substituting $$\vec{n}$$:

$$\langle 5, -1, -1 \rangle \cdot \langle x-1, y-2, z-3 \rangle = 0$$

This simplifies to:

$$5(x - 1) - 1(y - 2) - 1(z - 3) = 0$$

Expanding each term:

$$5(x - 1) = 5x - 5$$

$$-1(y - 2) = -y + 2$$

$$-1(z - 3) = -z + 3$$

Combining all:

$$5x - 5 - y + 2 - z + 3 = 0$$

Simplifying the constants:

$$5x - y - z - 5 + 2 + 3 = 0$$

$$5x - y - z + 0 = 0$$

So the plane equation is:

$$5x - y - z = 0$$

Now, check which option point satisfies this equation.

Option A: $$(1, -2, 5)$$

Substitute $$x = 1$$, $$y = -2$$, $$z = 5$$:

$$5(1) - (-2) - 5 = 5 + 2 - 5 = 2 \neq 0$$

Not satisfied.

Option B: $$(1, 0, 5)$$

Substitute $$x = 1$$, $$y = 0$$, $$z = 5$$:

$$5(1) - 0 - 5 = 5 - 0 - 5 = 0$$

Satisfied.

Option C: $$(0, 3, -5)$$

Substitute $$x = 0$$, $$y = 3$$, $$z = -5$$:

$$5(0) - 3 - (-5) = 0 - 3 + 5 = 2 \neq 0$$

Not satisfied.

Option D: $$(-1, -3, 0)$$

Substitute $$x = -1$$, $$y = -3$$, $$z = 0$$:

$$5(-1) - (-3) - 0 = -5 + 3 - 0 = -2 \neq 0$$

Not satisfied.

Only Option B satisfies the plane equation. Hence, the plane passes through the point $$(1, 0, 5)$$.

So, the answer is Option B.