If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:

${{ \textbf{(A)}\ \pm \frac{2}{3} \qquad\textbf{(B)}\ -\frac{2}{3}\qquad\textbf{(C)}\ -\frac{3}{2} \qquad\textbf{(D)}\ 6}\qquad\textbf{(E)}\ -6} $
1. To determine the value of \( a \) such that the graphs of the lines \( 2y + x + 3 = 0 \) and \( 3y + ax + 2 = 0 \) meet at right angles, we need to find the slopes of these lines and use the property that the slopes of perpendicular lines are negative reciprocals of each other.

2. First, we find the slope of the line \( 2y + x + 3 = 0 \). We can rewrite this equation in the slope-intercept form \( y = mx + b \):
   \[
   2y + x + 3 = 0 \implies 2y = -x - 3 \implies y = -\frac{1}{2}x - \frac{3}{2}
   \]
   Therefore, the slope \( m_1 \) of the first line is \( -\frac{1}{2} \).

3. Next, we find the slope of the line \( 3y + ax + 2 = 0 \). Similarly, we rewrite this equation in the slope-intercept form:
   \[
   3y + ax + 2 = 0 \implies 3y = -ax - 2 \implies y = -\frac{a}{3}x - \frac{2}{3}
   \]
   Therefore, the slope \( m_2 \) of the second line is \( -\frac{a}{3} \).

4. Since the lines are perpendicular, their slopes must be negative reciprocals of each other:
   \[
   m_1 \cdot m_2 = -1
   \]
   Substituting the slopes we found:
   \[
   \left(-\frac{1}{2}\right) \cdot \left(-\frac{a}{3}\right) = -1
   \]
   Simplifying this equation:
   \[
   \frac{a}{6} = -1 \implies a = -6
   \]

5. Therefore, the value of \( a \) is \( -6 \).

The final answer is \(\boxed{-6}\).