Find the value of $a$ so that the lines described by
\[\begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} a \\ -2 \\ 1 \end{pmatrix}\]and
\[\begin{pmatrix} 1 \\ -3/2 \\ -5 \end{pmatrix} + u \begin{pmatrix} 1 \\ 3/2 \\ 2 \end{pmatrix}\]are perpendicular.
To find the value of $a$ that makes the lines perpendicular, I need to use the fact that perpendicular lines have dot products equal to zero. The dot product of two lines can be computed by taking the dot product of their direction vectors. The direction vectors of the lines are $\begin{pmatrix} a \\ -2 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 3/2 \\ 2 \end{pmatrix}$. So I need to find $a$ such that $\begin{pmatrix} a \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3/2 \\ 2 \end{pmatrix} = 0$. To compute the dot product, I multiply the corresponding entries of the vectors and add them up. So I have $a \cdot 1 + (-2) \cdot (3/2) + 1 \cdot 2 = 0$. Simplifying, I get $a - 3 + 2 = 0$. Solving for $a$, I get $a = 1$.

