Distance Between Two Parallel Lines

To find the distance between two parallel lines, their coefficients of $$x$$ and $$y$$ must be identical. Let's rewrite the given equations:
$$\text{Line 1: } 3x - 4y + \sqrt{2}\sin\theta = 0$$
$$\text{Line 2: } 6x - 8y - 3\sqrt{2}\cos\theta = 0 \implies 3x - 4y - \frac{3\sqrt{2}}{2}\cos\theta = 0$$

Now, both lines match the standard form $$Ax + By + C = 0$$, where:
$$A = 3, \quad B = -4$$
$$C_1 = \sqrt{2}\sin\theta, \quad C_2 = -\frac{3\sqrt{2}}{2}\cos\theta$$

$$ \textbf{Step 2: Apply the parallel line distance formula} $$
The formula for the distance $d$ between two parallel lines is:
$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

Substitute the values into the formula:
$$d = \frac{\left| \sqrt{2}\sin\theta - \left(-\frac{3\sqrt{2}}{2}\cos\theta\right) \right|}{\sqrt{3^2 + (-4)^2}}$$
$$d = \frac{\left| \sqrt{2}\sin\theta + \frac{3\sqrt{2}}{2}\cos\theta \right|}{\sqrt{9 + 16}}$$
$$d = \frac{\sqrt{2}}{5} \left| \sin\theta + \frac{3}{2}\cos\theta \right|$$

$$ \textbf{Step 3: Maximize the trigonometric expression} $$
We know that the maximum value of any expression in the form $a\sin\theta + b\cos\theta$ is given by $$\sqrt{a^2 + b^2}$$.

Here, $$a = 1$$ and $$b = \frac{3}{2}$$. Let's calculate the maximum value:
$$\text{Maximum value of } \left( \sin\theta + \frac{3}{2}\cos\theta \right) = \sqrt{1^2 + \left(\frac{3}{2}\right)^2} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$

$$\textbf{Step 4: Calculate the maximum distance} $$
Substitute this maximum value back into the distance equation:
$$d_{\max} = \frac{\sqrt{2}}{5} \times \frac{\sqrt{13}}{2} = \frac{\sqrt{26}}{10}$$