So, for $\textbf{r}=\textbf{r}(t)$,
$\textbf{r}=(t+3)\textbf{i}+(2t-t^{2})\textbf{j}+(3t-t^{2})\textbf{k}$ = displacement
Therefore,
$\textbf{v}=\textbf{r}'=\textbf{i}+2\textbf{j}-2t\textbf{j}+3\textbf{k}-2t\textbf{k}$ = velocity
And,
$\textbf{a}=\textbf{r}''=-2\textbf{j}-2\textbf{k}$ = acceleration
To show orthogonality, I know $\textbf{a} \cdot \textbf{b}=0$, so for velocity and acceleration to be orthogonal (or perpendicular), $\textbf{v} \cdot \textbf{a}=0$.
$\textbf{v} \cdot \textbf{a}= (\textbf{i}+2\textbf{j}-2t\textbf{j}+3\textbf{k}-2t\textbf{k}) \cdot (-2\textbf{j}-2\textbf{k})=[\textbf{i}+(2+2t)\textbf{j}+(3-2t)\textbf{k}] \cdot (-2\textbf{j}-2\textbf{k})=(1)(0)\textbf{i}+(2+2t)(-2)\textbf{j}+(3-2t)(-2)\textbf{k}=-4\textbf{j}-4t\textbf{j}-6\textbf{k}+4t\textbf{k}=0$
$\implies 4t\textbf{j}-4t\textbf{k}=-4\textbf{j}-6\textbf{k}$
Obviously, there is no value for $t$ which satisfies the above, so I think I may have made a mistake!
Thanks.
As you wrote, $\textbf v=\textbf i+(2-2t)\cdot \textbf j + (3-2t)\cdot \textbf k$ and $\textbf a=-2\cdot \textbf j-2\cdot \textbf k$. The vector product and the solution when its zero is derived by $$\textbf v\cdot \textbf a =-2(2-2t)-2(3-2t)=0\implies t=\frac{5}{4}$$
In your solution, you should note that $\textbf j\cdot \textbf j = \textbf k\cdot \textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.