Let $$\theta_1$$ be the angle between two lines $$2x + 3y + c_1 = 0$$ and $$-x + 5y + c_2 = 0$$ and $$\theta_2$$ be the angle between two lines $$2x + 3y + c_1 = 0$$ and $$-x + 5y + c_3 = 0$$, where $$c_1, c_2, c_3$$ are any real numbers :
Statement-1: If $$c_2$$ and $$c_3$$ are proportional, then $$\theta_1 = \theta_2$$.
Statement-2: $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$.

To solve this problem, we need to evaluate two statements about the angles between given lines. Recall that the angle between two lines depends only on their slopes, not on the constant terms in their equations. The slope of a line given by $$ax + by + c = 0$$ is $$m = -\frac{a}{b}$$.

First, consider the line $$2x + 3y + c_1 = 0$$. Its slope is $$m_1 = -\frac{2}{3}$$. Now, the line $$-x + 5y + c_2 = 0$$ has slope $$m_2 = -\frac{-1}{5} = \frac{1}{5}$$. Similarly, the line $$-x + 5y + c_3 = 0$$ has slope $$m_3 = -\frac{-1}{5} = \frac{1}{5}$$. Notice that both lines $$-x + 5y + c_2 = 0$$ and $$-x + 5y + c_3 = 0$$ have the same slope $$\frac{1}{5}$$, regardless of the values of $$c_2$$ and $$c_3$$. This is because the constants $$c_2$$ and $$c_3$$ only shift the lines parallel to themselves without changing their slopes.

The angle $$\theta_1$$ is between the lines $$2x + 3y + c_1 = 0$$ (slope $$m_1 = -\frac{2}{3}$$) and $$-x + 5y + c_2 = 0$$ (slope $$m_2 = \frac{1}{5}$$). The formula for the tangent of the angle $$\theta$$ between two lines with slopes $$m_1$$ and $$m_2$$ is:

$$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substituting the slopes:

$$\tan \theta_1 = \left| \frac{-\frac{2}{3} - \frac{1}{5}}{1 + \left(-\frac{2}{3}\right)\left(\frac{1}{5}\right)} \right| = \left| \frac{ -\frac{10}{15} - \frac{3}{15} }{1 - \frac{2}{15}} \right| = \left| \frac{ -\frac{13}{15} }{ \frac{15}{15} - \frac{2}{15} } \right| = \left| \frac{ -\frac{13}{15} }{ \frac{13}{15} } \right| = \left| -1 \right| = 1$$

Similarly, the angle $$\theta_2$$ is between the lines $$2x + 3y + c_1 = 0$$ (slope $$m_1 = -\frac{2}{3}$$) and $$-x + 5y + c_3 = 0$$ (slope $$m_3 = \frac{1}{5}$$). Since $$m_3 = m_2 = \frac{1}{5}$$, the calculation is identical:

$$\tan \theta_2 = \left| \frac{-\frac{2}{3} - \frac{1}{5}}{1 + \left(-\frac{2}{3}\right)\left(\frac{1}{5}\right)} \right| = \left| \frac{ -\frac{10}{15} - \frac{3}{15} }{1 - \frac{2}{15}} \right| = \left| \frac{ -\frac{13}{15} }{ \frac{13}{15} } \right| = \left| -1 \right| = 1$$

Thus, $$\tan \theta_1 = \tan \theta_2 = 1$$, so $$\theta_1 = \theta_2 = 45^\circ$$ (since the angle between lines is acute and $$\tan^{-1}(1) = 45^\circ$$). Importantly, this result holds for any real numbers $$c_1$$, $$c_2$$, and $$c_3$$ because the constants do not affect the slopes.

Now, evaluate Statement-2: $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$. As shown above, since the slopes are identical and independent of $$c_2$$ and $$c_3$$, $$\theta_1$$ and $$\theta_2$$ are always equal. Thus, Statement-2 is true.

Statement-1 states that if $$c_2$$ and $$c_3$$ are proportional (i.e., $$c_3 = k c_2$$ for some constant $$k$$), then $$\theta_1 = \theta_2$$. Since we have established that $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$, this includes the case where $$c_2$$ and $$c_3$$ are proportional. Therefore, Statement-1 is also true.

However, Statement-2 provides the complete reason: $$\theta_1 = \theta_2$$ because the slopes are fixed regardless of $$c_2$$ and $$c_3$$. Statement-1 only specifies a subset of cases (proportional constants), but the equality holds universally. Thus, Statement-2 explains why Statement-1 is true, as the condition in Statement-1 is sufficient but not necessary, and the underlying reason is the slope invariance captured by Statement-2.

Hence, both statements are true, and Statement-2 is a correct explanation for Statement-1.

So, the answer is Option A.