If the line 2x - 3y + 17 = 0 is perpendicular to the line L that passes through the points P = (7, 17) and Q = (15, m) then the equation of the line L is:

Given that the line 2x - 3y + 17 = 0 is perpendicular to the line L passing through points P = (7, 17) and Q = (15, m), we start by finding the slope of the given line.

To find the slope of the line 2x - 3y + 17 = 0, we rewrite it in slope-intercept form:

2x - 3y + 17 = 0 \implies 3y = 2x + 17 \implies y = \frac{2}{3}x + \frac{17}{3}

The slope (m) of this line is:

m_1 = \frac{2}{3}

Since the line is perpendicular to line L, the slope of line L is the negative reciprocal of \frac{2}{3}, which is:

m_2 = -\frac{3}{2}

The line L passes through points P = (7, 17) and Q = (15, m). Using the point-slope form of the equation of a line, we know that the line passing through P with slope m_2 is given by:

y - y_1 = m_2 (x - x_1)

Substituting P = (7, 17) and m_2 = -\frac{3}{2}, we get:

y - 17 = -\frac{3}{2} (x - 7)

Simplify and put it in slope-intercept form:

y - 17 = -\frac{3}{2}x + \frac{21}{2}

y = -\frac{3}{2}x + \frac{21}{2} + 17

y = -\frac{3}{2}x + \frac{21}{2} + \frac{34}{2}

y = -\frac{3}{2}x + \frac{55}{2}

Finally, convert it to the standard form:

2y = -3x + 55

3x + 2y = 55

Thus, the equation of the line L is:

\boxed{3x + 2y = 55}