Answer
The tangent vector's head always lie in the plane $z=-1$. Therefore, there is no way it can head to the $x$-axis. Hence, the pilot cannot hit any target on the $x$-axis.
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 ${\bf{r}}\left( t \right) = \left( {t - {t^3},12 - {t^2},3 - t} \right)$ we get ${\bf{r}}'\left( t \right) = \left( {1 - 3{t^2}, - 2t, - 1} \right)$. Thus, the tangent vector is ${\bf{r}}'\left( t \right) = \left( {1 - 3{t^2}, - 2t, - 1} \right)$.
Recall from Exercise 57, ${\bf{r}}\left( t \right)$ is the position vector of the fighter plane which can shoot a laser beam straight ahead. However, the tangent vector ${\bf{r}}'\left( t \right) = \left( {1 - 3{t^2}, - 2t, - 1} \right)$ implies that its head always lie in the plane $z=-1$. Therefore, there is no way it can head to the $x$-axis. Hence, the pilot cannot hit any target on the $x$-axis.
