(a) If p and q are perpendicular (or orthogonal), then their dot product is 0:
[tex]\mathbf p^\top\mathbf q = \begin{pmatrix}3&t\end{pmatrix}\begin{pmatrix}-2\\1\end{pmatrix} = 3\times(-2)+t\times1 = t-6 = 0 \implies \boxed{t=6}[/tex]
(b) and (c) can be answered in the same way.
(b)
[tex]\begin{pmatrix}t&t+2\end{pmatrix}\begin{pmatrix}3\\-4\end{pmatrix} = 3t-4(t+2) = -t-8 = 0 \implies \boxed{t=-8}[/tex]
(c)
[tex]\begin{pmatrix}t&t+2\end{pmatrix}\begin{pmatrix}2-3t\\t\end{pmatrix} = t(2-3t)+(t+2)t = 4t-2t^2 = 2t(2-t)=0 \implies \boxed{t=0\text{ or }t=2}[/tex]
