Answer
$x=-\pi-t,\ \ \ y=\pi+t$, $\ \ $ $z =-\pi t$
Work Step by Step
Work done in Geogebra CAS (free online).
From the parametric equations, for the point $(-\pi,\pi,0)$, t is $\pi.$.
CAS entry: k:=pi
Next, define r(t) and r'(t)
CAS entry: $r(t):=(t\cos t,t,t\sin t)$
CAS entry: $r1(t)=$Derivative( r(t))
To plot the point $(-\pi,\pi,0)$:
CAS entry: $A=r(k)$
To find the tangent vector at $(-\pi,\pi,0)$:
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 $(-\pi,\pi,0)$ and is parallel to the vector $\langle-1,\ 1, -\pi\rangle$.
Its parametric equations are (see sec 12-5):
$x=-\pi-t,\ \ \ y=\pi+t$, $\ \ $ $z =-\pi t$
