Given that the vectors (t, 4, 2t + 1) and (t + 2, 1 - t, -1) are perpendicular, find the possible values of the constant t.

Let the given vectors be $\mathbf{a}$ and $\mathbf{b}$.

$$\mathbf{a} = \begin{pmatrix} t \\ 4 \\ 2t+1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} t+2 \\ 1-t \\ -1 \end{pmatrix}$$

The dot product $\mathbf{a} \cdot \mathbf{b}$ is calculated as:

$$t(t+2) + 4(1-t) + (2t+1)(-1) = 0$$

$$t^2 + (2t - 4t - 2t) + (4 - 1) = 0$$ $$t^2 - 4t + 3 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.

$$(t-1)(t-3) = 0$$

$$t-1 = 0 \implies t = 1$$ $$t-3 = 0 \implies t = 3$$

The possible values for $t$ are 1 and 3.

The possible values of the constant $t$ are $\boxed{\text{1 and 3}}$.