If $$Z_1 \neq 0$$ and $$Z_2$$ be two complex numbers such that $$\frac{Z_2}{Z_1}$$ is a purely imaginary number, then $$\left|\frac{2Z_1 + 3Z_2}{2Z_1 - 3Z_2}\right|$$ is equal to:

Given that $$ z_1 \neq 0 $$ and $$ z_2 $$ are complex numbers such that $$ \frac{z_2}{z_1} $$ is purely imaginary. This means $$ \frac{z_2}{z_1} = ki $$ for some real number $$ k $$. Therefore, we can write $$ z_2 = ki z_1 $$.

We need to find $$ \left| \frac{2z_1 + 3z_2}{2z_1 - 3z_2} \right| $$. Substitute $$ z_2 = ki z_1 $$ into the expression:

The numerator becomes $$ 2z_1 + 3z_2 = 2z_1 + 3(ki z_1) = z_1 (2 + 3ki) $$.

The denominator becomes $$ 2z_1 - 3z_2 = 2z_1 - 3(ki z_1) = z_1 (2 - 3ki) $$.

So the fraction is:

$$ \frac{2z_1 + 3z_2}{2z_1 - 3z_2} = \frac{z_1 (2 + 3ki)}{z_1 (2 - 3ki)} = \frac{2 + 3ki}{2 - 3ki} $$

Since $$ z_1 \neq 0 $$, we cancel $$ z_1 $$. Now, we find the modulus of this complex fraction. The modulus of a quotient is the quotient of the moduli:

$$ \left| \frac{2 + 3ki}{2 - 3ki} \right| = \frac{|2 + 3ki|}{|2 - 3ki|} $$

The modulus of $$ 2 + 3ki $$ is $$ \sqrt{2^2 + (3k)^2} = \sqrt{4 + 9k^2} $$.

The modulus of $$ 2 - 3ki $$ is $$ \sqrt{2^2 + (-3k)^2} = \sqrt{4 + 9k^2} $$.

Both moduli are equal, so:

$$ \frac{|2 + 3ki|}{|2 - 3ki|} = \frac{\sqrt{4 + 9k^2}}{\sqrt{4 + 9k^2}} = 1 $$

Thus, the modulus is 1.

Alternatively, we can compute the modulus by rationalizing the fraction. Let $$ t = \frac{2 + 3ki}{2 - 3ki} $$. Multiply numerator and denominator by the conjugate of the denominator, $$ 2 + 3ki $$:

$$ t = \frac{(2 + 3ki)(2 + 3ki)}{(2 - 3ki)(2 + 3ki)} = \frac{(2 + 3ki)^2}{2^2 - (3ki)^2} = \frac{(2 + 3ki)^2}{4 - 9k^2 i^2} $$

Since $$ i^2 = -1 $$,

$$ t = \frac{(2 + 3ki)^2}{4 - 9k^2 (-1)} = \frac{(2 + 3ki)^2}{4 + 9k^2} $$

Expand the numerator: $$ (2 + 3ki)^2 = 4 + 2 \cdot 2 \cdot 3ki + (3ki)^2 = 4 + 12ki + 9k^2 i^2 = 4 + 12ki - 9k^2 = (4 - 9k^2) + 12ki $$.

So,

$$ t = \frac{(4 - 9k^2) + 12ki}{4 + 9k^2} = \frac{4 - 9k^2}{4 + 9k^2} + i \frac{12k}{4 + 9k^2} $$

The modulus is:

$$ |t| = \sqrt{ \left( \frac{4 - 9k^2}{4 + 9k^2} \right)^2 + \left( \frac{12k}{4 + 9k^2} \right)^2 } $$

Compute the expression inside the square root:

$$ \left( \frac{4 - 9k^2}{4 + 9k^2} \right)^2 + \left( \frac{12k}{4 + 9k^2} \right)^2 = \frac{(4 - 9k^2)^2 + (12k)^2}{(4 + 9k^2)^2} $$

Expand the numerator: $$ (4 - 9k^2)^2 = 16 - 72k^2 + 81k^4 $$, and $$ (12k)^2 = 144k^2 $$, so:

$$ 16 - 72k^2 + 81k^4 + 144k^2 = 16 + 72k^2 + 81k^4 $$

The denominator is $$ (4 + 9k^2)^2 = 16 + 72k^2 + 81k^4 $$. Thus,

$$ \frac{16 + 72k^2 + 81k^4}{16 + 72k^2 + 81k^4} = 1 $$

So, $$ |t| = \sqrt{1} = 1 $$.

Both methods confirm that the modulus is 1. Hence, the expression $$ \left| \frac{2z_1 + 3z_2}{2z_1 - 3z_2} \right| = 1 $$.

Comparing with the options: A is 2, B is 5, C is 3, D is 1. Therefore, the correct option is D.

Hence, the correct answer is Option D.