To find the perpendicular distance from the point $$P(2,-10,1)$$ to the line given by

$$\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2},$$

follow these steps:

Parametrize the Line:

From the given symmetric equations, set the common parameter as $$t$$:

$$x = 1 + 2t$$

$$y = -2 - t$$

$$z = -3 + 2t$$

This shows that:

A point on the line is $$A(1,-2,-3)$$ (when $$t=0$$).

The direction vector is $$\vec{d} = \langle 2, -1, 2 \rangle.$$

Determine the Vector from Point A to P:

Calculate

$$\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.$$

Compute the Cross Product:

The formula for the perpendicular distance from a point to a line in 3D is:

$$d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.$$

First, find the cross product $$\vec{AP} \times \vec{d}$$:

$$ \vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big). $$

Evaluate each component:

First component: $$(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.$$

Second component: $$ (4 \times 2) - (1 \times 2) = 8 - 2 = 6.$$

Third component: $$ (1 \times -1) - (-8 \times 2) = -1 + 16 = 15.$$

So,

$$\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.$$

Calculate the Magnitudes:

For the cross product:

$$ \|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}. $$

For the direction vector:

$$ \|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. $$

Compute the Distance:

Substitute the magnitudes into the distance formula:

$$ d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}. $$

Thus, the perpendicular distance from the point $$P(2,-10,1)$$ to the line is $$3\sqrt{5}$$.

Comparing with the options given, the correct answer is:

Option C: $$3\sqrt{5}$$.