Hi

As far as I remember the curvature is the norm of vectorr"(t)

You just have to calculate the coordinates ofr"(t) and then its norm

One vector of the tangent line isr'(t=-2pi/6)

As soon as you get its coordinates (a,b,c) you can have the parametric equation of the tangent line

x(t) = at + cos(-2pi/6)

y(t) = bt + sin(-2pi/6)

z(t) = ct + (-2pi/6)