If the distance between planes, $$4x - 2y - 4z + 1 = 0$$ and $$4x - 2y - 4z + d = 0$$ is 7, then d is:

The given planes are $$4x - 2y - 4z + 1 = 0$$ and $$4x - 2y - 4z + d = 0$$. These planes are parallel because they have the same normal vector $$\langle 4, -2, -4 \rangle$$. The distance between two parallel planes $$ax + by + cz + d_1 = 0$$ and $$ax + by + cz + d_2 = 0$$ is given by the formula:

$$\text{Distance} = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$$

For the first plane, $$4x - 2y - 4z + 1 = 0$$, we have $$d_1 = 1$$. For the second plane, $$4x - 2y - 4z + d = 0$$, we have $$d_2 = d$$. The coefficients are $$a = 4$$, $$b = -2$$, and $$c = -4$$.

First, compute the denominator $$\sqrt{a^2 + b^2 + c^2}$$:

$$a^2 = 4^2 = 16$$

$$b^2 = (-2)^2 = 4$$

$$c^2 = (-4)^2 = 16$$

Adding these: $$16 + 4 + 16 = 36$$

So, $$\sqrt{36} = 6$$

The distance is given as 7, so:

$$7 = \frac{|1 - d|}{6}$$

Multiply both sides by 6 to isolate the absolute value:

$$7 \times 6 = |1 - d|$$

$$42 = |1 - d|$$

The absolute value equation $$|1 - d| = 42$$ means that $$1 - d$$ could be 42 or $$1 - d$$ could be -42. We solve both cases.

Case 1: $$1 - d = 42$$

Subtract 1 from both sides:

$$-d = 42 - 1$$

$$-d = 41$$

Multiply both sides by -1:

$$d = -41$$

Case 2: $$1 - d = -42$$

Subtract 1 from both sides:

$$-d = -42 - 1$$

$$-d = -43$$

Multiply both sides by -1:

$$d = 43$$

Therefore, the possible values of $$d$$ are $$-41$$ and $$43$$.

Comparing with the options:

A. 41 or $$-42$$

B. 42 or $$-43$$

C. $$-41$$ or 43

D. $$-42$$ or 44

The values $$-41$$ and $$43$$ match option C.

Hence, the correct answer is Option C.