Answer
There is precisely one time $t$ at which the pilot can hit a target located at the origin, which is at $t=3$.
Work Step by Step
Recall that the derivative vector ${\bf{r}}'\left( {{t_0}} \right)$ points in the direction tangent to the path traced by ${\bf{r}}\left( t \right)$ at $t = {t_0}$. For the curve ${\bf{r}}\left( t \right) = \left( {5 - t,21 - {t^2},3 - {t^3}/27} \right)$ we get ${\bf{r}}'\left( t \right) = \left( { - 1, - 2t, - {t^2}/9} \right)$. Thus, the tangent vector is ${\bf{r}}'\left( t \right) = \left( { - 1, - 2t, - {t^2}/9} \right)$.
Since ${\bf{r}}\left( t \right)$ is the position vector of the fighter plane, which can shoot a laser beam straight ahead, it can hit a target located at the origin if ${\bf{r}}\left( t \right)$ and ${\bf{r}}'\left( t \right)$ are parallel and point in the opposite direction. Thus,
${\bf{r}}\left( t \right) = - \lambda {\bf{r}}'\left( t \right)$,
where $\lambda > 0$ is a scalar. So,
$\left( {5 - t,21 - {t^2},3 - {t^3}/27} \right) = - \lambda \left( { - 1, - 2t, - {t^2}/9} \right)$
We have a system of equations:
$5 - t = \lambda $, ${\ \ }$ $21 - {t^2} = 2\lambda t$, ${\ \ }$ $3 - \frac{{{t^3}}}{{27}} = \lambda \frac{{{t^2}}}{9}$
From the first equation, we obtain $\lambda = 5 - t$. Substituting it in the second equation gives
$21 - {t^2} = 2t\left( {5 - t} \right)$
$21 - {t^2} + 2{t^2} - 10t = 0$
${t^2} - 10t + 21 = 0$
$\left( {t - 3} \right)\left( {t - 7} \right) = 0$
The solutions are ${t_1} = 3$ and ${t_2} = 7$. The corresponding values for $\lambda$ are ${\lambda _1} = 2$ and ${\lambda _2} = - 2$. However, since ${\bf{r}}\left( t \right)$ and ${\bf{r}}'\left( t \right)$ must be in the opposite direction, there is only one solution in this case, that is ${t_1} = 3$ and ${\lambda _1} = 2$. Substituting this solution in the third equation verifies that it also satisfies the third equation. Hence, there is precisely one time $t$ at which the pilot can hit a target located at the origin, which is at $t=3$.
