A plane containing the point $$(3, 2, 0)$$ and the line $$\frac{x-1}{1} = \frac{y-2}{5} = \frac{z-3}{4}$$ also contains the point

To solve this problem, we need to find which of the given points lies on the plane that contains the point $$(3, 2, 0)$$ and the line $$\frac{x-1}{1} = \frac{y-2}{5} = \frac{z-3}{4}$$. First, we express the line in parametric form. Let the common ratio be $$t$$, so:

$$x - 1 = t \quad \Rightarrow \quad x = t + 1$$

$$y - 2 = 5t \quad \Rightarrow \quad y = 5t + 2$$

$$z - 3 = 4t \quad \Rightarrow \quad z = 4t + 3$$

Thus, any point on the line is $$(t + 1, 5t + 2, 4t + 3)$$. Since the plane contains this entire line and the point $$(3, 2, 0)$$, we can find three points on the plane by choosing specific values of $$t$$. For $$t = 0$$, we get point $$A(1, 2, 3)$$. For $$t = 1$$, we get point $$B(2, 7, 7)$$. The given point is $$C(3, 2, 0)$$.

Now, we find the equation of the plane passing through these three points: $$A(1, 2, 3)$$, $$B(2, 7, 7)$$, and $$C(3, 2, 0)$$. The general equation of a plane is $$ax + by + cz = d$$. To find the coefficients, we compute two vectors in the plane:

Vector $$\overrightarrow{AC} = C - A = (3 - 1, 2 - 2, 0 - 3) = (2, 0, -3)$$

Vector $$\overrightarrow{AB} = B - A = (2 - 1, 7 - 2, 7 - 3) = (1, 5, 4)$$

The normal vector $$\vec{N}$$ to the plane is the cross product of $$\overrightarrow{AC}$$ and $$\overrightarrow{AB}$$:

$$\vec{N} = \overrightarrow{AC} \times \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -3 \\ 1 & 5 & 4 \end{vmatrix}$$

Expanding the determinant:

$$\hat{i} \left( (0)(4) - (-3)(5) \right) - \hat{j} \left( (2)(4) - (-3)(1) \right) + \hat{k} \left( (2)(5) - (0)(1) \right)$$

$$= \hat{i} (0 - (-15)) - \hat{j} (8 - (-3)) + \hat{k} (10 - 0)$$

$$= \hat{i} (15) - \hat{j} (11) + \hat{k} (10)$$

$$= (15, -11, 10)$$

So, the normal vector is $$(15, -11, 10)$$. Using point $$A(1, 2, 3)$$, the plane equation is:

$$15(x - 1) - 11(y - 2) + 10(z - 3) = 0$$

Expanding and simplifying:

$$15x - 15 - 11y + 22 + 10z - 30 = 0$$

$$15x - 11y + 10z - 23 = 0$$

$$15x - 11y + 10z = 23$$

Now, we check which of the given options satisfies this plane equation.

Option A: $$(0, 7, -10)$$

Substitute $$x = 0$$, $$y = 7$$, $$z = -10$$:

$$15(0) - 11(7) + 10(-10) = 0 - 77 - 100 = -177 \neq 23$$

Not on the plane.

Option B: $$(0, 7, 10)$$

Substitute $$x = 0$$, $$y = 7$$, $$z = 10$$:

$$15(0) - 11(7) + 10(10) = 0 - 77 + 100 = 23 = 23$$

On the plane.

Option C: $$(0, 3, 1)$$

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

$$15(0) - 11(3) + 10(1) = 0 - 33 + 10 = -23 \neq 23$$

Not on the plane.

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

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

$$15(0) - 11(-3) + 10(1) = 0 + 33 + 10 = 43 \neq 23$$

Not on the plane.

Only option B satisfies the plane equation. To verify, we check a point on the line, say $$t=0$$: $$(1, 2, 3)$$:

$$15(1) - 11(2) + 10(3) = 15 - 22 + 30 = 23$$

And the given point $$(3, 2, 0)$$:

$$15(3) - 11(2) + 10(0) = 45 - 22 = 23$$

Both satisfy, confirming the plane equation is correct.

Hence, the correct answer is Option B.