For two lines to be perpendicular, the product of their slopes must be \( -1 \). We begin by finding the slopes of the given lines.
1. First line equation: \( 3x - 2y = 5 \)
Rearrange into slope-intercept form \( y = mx + c \):
\[
3x - 2y = 5 \quad \Rightarrow \quad -2y = -3x + 5 \quad \Rightarrow \quad y = \frac{3}{2}x - \frac{5}{2}
\]
The slope of the first line is \( m_1 = \frac{3}{2} \).
2. Second line equation: \( 2x + ky + 7 = 0 \)
Rearrange into slope-intercept form:
\[
2x + ky + 7 = 0 \quad \Rightarrow \quad ky = -2x - 7 \quad \Rightarrow \quad y = -\frac{2}{k}x - \frac{7}{k}
\]
The slope of the second line is \( m_2 = -\frac{2}{k} \).
For the lines to be perpendicular:
\[
m_1 \times m_2 = -1
\]
Substitute the values of \( m_1 \) and \( m_2 \):
\[
\frac{3}{2} \times \left( -\frac{2}{k} \right) = -1 \quad \Rightarrow \quad -\frac{6}{2k} = -1 \quad \Rightarrow \quad \frac{6}{2k} = 1 \quad \Rightarrow \quad k = 3
\]
Thus, the solution is \( k = 3 \).