Answer
The two curves intersect. The point of intersection is $\left( {2,4,3} \right)$.
Work Step by Step
The two curves intersect if there exist parameter values $t$ and $s$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( s \right)$. So,
$x = t = \sqrt s $, ${\ \ }$ $y = {t^2} = s$, ${\ \ }$ $z = t + 1 = s - 1$
From the third equation we obtain $s=t+2$. Substituting it in the second equation gives
${t^2} = t + 2$
${t^2} - t - 2 = 0$
This is a quadratic equation whose solutions are
${t_{1,2}} = \frac{{1 \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\cdot1\cdot\left( { - 2} \right)} }}{{2\cdot1}} = \frac{{1 \pm \sqrt 9 }}{2}$
${t_{1,2}} = \frac{1}{2} \pm \frac{3}{2}$
So, ${t_1} = 2$ and ${t_2} = - 1$. However, the parametrization of ${{\bf{r}}_2}\left( s \right)$ requires that $t>0$, that is, the $x$-component be positive. Therefore, we only have one solution, that is ${t_1} = 2$. The corresponding value of $s$ is $s=4$.
We obtain the point of intersection by substituting ${t_1} = 2$ in ${{\bf{r}}_1}\left( t \right)$:
${{\bf{r}}_1}\left( 2 \right) = \left( {2,4,3} \right)$
Thus, the point of intersection is $\left( {2,4,3} \right)$.
