The equation of a plane through the line of intersection of the planes $$x + 2y = 3$$, $$y - 2z + 1 = 0$$, and perpendicular to the first plane is :

The given planes are $$x + 2y = 3$$ and $$y - 2z + 1 = 0$$. To find the equation of a plane passing through their line of intersection and perpendicular to the first plane, we use the family of planes passing through the line of intersection.

The general equation for such a family is derived by combining the two plane equations with a parameter $$k$$. First, write both planes in standard form:

Plane 1: $$x + 2y + 0z - 3 = 0$$

Plane 2: $$0x + y - 2z + 1 = 0$$

The family of planes is:

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

Expanding this:

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

$$x + (2 + k)y - 2k z + (k - 3) = 0$$

This plane must be perpendicular to the first plane $$x + 2y - 3 = 0$$. The normal vector of the first plane is $$\vec{n_1} = (1, 2, 0)$$. The normal vector of the new plane is $$\vec{n_2} = (1, 2 + k, -2k)$$.

For perpendicular planes, the dot product of the normal vectors is zero:

$$\vec{n_1} \cdot \vec{n_2} = 0$$

$$(1)(1) + (2)(2 + k) + (0)(-2k) = 0$$

$$1 + 4 + 2k = 0$$

$$5 + 2k = 0$$

$$2k = -5$$

$$k = -\frac{5}{2}$$

Substitute $$k = -\frac{5}{2}$$ back into the family equation:

$$x + \left(2 + \left(-\frac{5}{2}\right)\right)y - 2\left(-\frac{5}{2}\right)z + \left(-\frac{5}{2} - 3\right) = 0$$

$$x + \left(2 - \frac{5}{2}\right)y + 5z + \left(-\frac{5}{2} - \frac{6}{2}\right) = 0$$

$$x + \left(-\frac{1}{2}\right)y + 5z + \left(-\frac{11}{2}\right) = 0$$

To eliminate fractions, multiply the entire equation by 2:

$$2x - y + 10z - 11 = 0$$

$$2x - y + 10z = 11$$

Comparing with the options, this matches option C.

Hence, the correct answer is Option C.