The distance of the point (1, 0, 2) from the point of intersection of the line $$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$$ and the plane $$x - y + z = 16$$, is

To find the distance from the point (1, 0, 2) to the intersection point of the line $$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$$ and the plane $$x - y + z = 16$$, we first need to determine the point where the line intersects the plane.

The symmetric equations of the line can be parameterized by setting $$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = t$$, where $$t$$ is a parameter. This gives:

$$x = 2 + 3t$$

$$y = -1 + 4t$$

$$z = 2 + 12t$$

Since this point lies on the plane $$x - y + z = 16$$, substitute the parametric equations into the plane equation:

$$(2 + 3t) - (-1 + 4t) + (2 + 12t) = 16$$

Simplify the expression inside. Note that $$-(-1 + 4t) = +1 - 4t$$, so:

$$2 + 3t + 1 - 4t + 2 + 12t = 16$$

Combine the constant terms: $$2 + 1 + 2 = 5$$. Combine the $$t$$ terms: $$3t - 4t + 12t = (3 - 4 + 12)t = 11t$$. This gives:

$$5 + 11t = 16$$

Solve for $$t$$ by subtracting 5 from both sides:

$$11t = 11$$

Divide both sides by 11:

$$t = 1$$

Substitute $$t = 1$$ back into the parametric equations to find the intersection point:

$$x = 2 + 3(1) = 2 + 3 = 5$$

$$y = -1 + 4(1) = -1 + 4 = 3$$

$$z = 2 + 12(1) = 2 + 12 = 14$$

Thus, the point of intersection is (5, 3, 14).

Now, calculate the distance from the point (1, 0, 2) to (5, 3, 14) using the distance formula in 3D space:

$$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Substitute the coordinates: $$x_1 = 1$$, $$y_1 = 0$$, $$z_1 = 2$$, $$x_2 = 5$$, $$y_2 = 3$$, $$z_2 = 14$$.

Compute the differences:

$$x_2 - x_1 = 5 - 1 = 4$$

$$y_2 - y_1 = 3 - 0 = 3$$

$$z_2 - z_1 = 14 - 2 = 12$$

Square each difference:

$$4^2 = 16$$

$$3^2 = 9$$

$$12^2 = 144$$

Sum the squares:

$$16 + 9 + 144 = 169$$

Take the square root:

$$\sqrt{169} = 13$$

Hence, the distance is 13. Comparing with the options, A is 13.

Hence, the correct answer is Option A.