Find the value of $a$ so that the lines described by
\[\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix} + t \begin{pmatrix} 2 \\ a \\ 4 \end{pmatrix}\]and
\[\begin{pmatrix} -7 \\ -3 \\ 11 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix}\]are perpendicular.
To find the value of $a$, I need to use the fact that perpendicular lines have dot products equal to zero. The direction vectors are \(\begin{pmatrix} 2 \\ a \\ 4 \end{pmatrix}\) and \(\begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix}\). The dot product is \((2)(-1) + (a)(4) + (4)(2)\). So I have the equation \(-(2) + (a)(4) + (4)(2) = 0\). Simplifying, I get \(4a + 6 = 0\). Solving for $a$, I get \(a = -\frac{3}{2}\). 