Answer
$x = 2 + 2t, y = 4 + 2t, z = 1 + t$
Work Step by Step
Given: $r(t) = (t^{2} + 1)i + (4\sqrt t)j + e^{t^{2}-t}k$
In order to find tangent vector $r'(t)$ we will have to take the derivative of $r(t)$.
Thus, $r'(t) = 2ti + \frac{2}{\sqrt t}j + e^{t^{2}-t}(2t-1)k$
Setting a component of $r(t)$ equal to the corresponding given point $(2,4,1)$ we get that $t=1$.
$r'(1) = 2(1)i + \frac{2}{\sqrt 1}j + e^{1^{2}-1}(2*1-1)k = 2i + 2j + k$
Equation of the tangent line passing through the point $(2,4,1)$ is: $$r(t)=<2+2t,4+2t,1+t>$$
Therefore, the parametric equations are: $$x = 2 + 2t, y = 4 + 2t, z = 1 + t$$
