Answer
$x=\sqrt{3}-t,\ \ \ y=1+\sqrt{3}t$, $\ \ $ $z =2-4\sqrt{3}t$
Work Step by Step
Work done in Geogebra CAS (free online).
From the parametric equations, for the point ($\sqrt{3},1,2$), t is $\pi/6.$.
CAS entry: k:=pi/6
Next, define r(t) and r'(t)
CAS entry: $r(t):=(2\cos t,2\sin t,4\cos 2t)$
CAS entry: $r1(t)=$Derivative( r(t))
To plot the point ($\sqrt{3},1,2$):
CAS entry: $A=r(k)$
To find the tangent vector at ($\sqrt{3},1,2$):
CAS entry: $u:=r1(k)$
We now use the point-vector facility of geogebra to draw the vector u from point A.
Then, the tangent line.
CAS entry: Line(A,u)
The tangent line contains ($\sqrt{3},1,2$) and is parallel to the vector $\langle-1,\ \sqrt{3}, -4\sqrt{3}\rangle$.
Its parametric equations are (see sec 12-5)
$x=\sqrt{3}-t,\ \ \ y=1+\sqrt{3}t$, $\ \ $ $z =2-4\sqrt{3}t$
