If the straight line $$2x - 3y + 17 = 0$$ is perpendicular to the line passing through the points $$(7, 17)$$ and $$(15, \beta)$$, then $$\beta$$ equals:

We have the first straight line $$2x - 3y + 17 = 0$$. In order to read its slope easily, we bring it to the slope-intercept form $$y = mx + c$$.

Starting with $$2x - 3y + 17 = 0,$$ we isolate the $$y$$-term:

$$-3y = -2x - 17.$$

Dividing every term by $$-3$$ gives

$$y = \frac{2}{3}x + \frac{17}{3}.$$

So the slope of this line is $$m_1 = \frac{2}{3}.$$

Now, two lines are perpendicular when the product of their slopes equals $$-1$$. Hence, if the slope of the required second line is called $$m_2$$, then

$$m_1 \, m_2 = -1 \quad\Longrightarrow\quad \frac{2}{3}\, m_2 = -1.$$

Solving for $$m_2$$:

$$m_2 = -\frac{3}{2}.$$

Next, the second line passes through the two given points $$(7,\,17)$$ and $$(15,\,\beta)$$. By the two-point slope formula, the slope of this line is

$$m_2 = \frac{\beta - 17}{15 - 7}.$$

Simplifying the denominator:

$$m_2 = \frac{\beta - 17}{8}.$$

We have already found that $$m_2 = -\dfrac{3}{2}$$, so we equate the two expressions for the same slope:

$$\frac{\beta - 17}{8} = -\frac{3}{2}.$$

To clear the fractions, we multiply both sides by $$8$$:

$$\beta - 17 = 8 \left(-\frac{3}{2}\right).$$

Calculating the right-hand side:

$$\beta - 17 = -12.$$

Finally, adding $$17$$ to both sides yields

$$\beta = -12 + 17 = 5.$$

Therefore, $$\beta$$ = 5.$$

Hence, the correct answer is Option 3.