As Mark points out, unit normal and principal normal is just the same thing.
Now we have:
Let $ v(t) = \frac{1}{\sqrt{1 + 4t^2 + 9t^4}} $
Tangent is $ (1, 2t, 3t^2) v $.
Normal is $ (0, 2, 6t) v(t) + (1, 2t, 3t^2) v'(t) $
$ v'(t) = -(1 + 4t^2 + 9t^4)^{-\frac{3}{2}} (4t + 18t^3) $
$ v(1) = \frac{1}{\sqrt{14}} $
$ v'(1) = -{14}^{-\frac{3}{2}} (22) = -\frac{11}{7\sqrt{14}} $
Unit tangent at $ t = 1 $ is $ T = \frac{1}{\sqrt{14}}(1, 2, 3) $
Unit normal at $ t = 1 $ is $ N = \frac{1}{\sqrt{266}}(-11, -8, 9) $
Unit binormal at $ t = 1 $ is $ T \times N = \frac{1}{\sqrt{19}}(3, -3, 1) $.
Together these form the Frenet frame for the curve.
Whenever we have $3$ orthogonal vectors in $\mathbb{R}^3$, they form a basis if and only if each of them is non-null. But you are assuming that $\mathbf{T}(s)\neq0$, $\bigl\lVert\mathbf{N}(s)\bigr\rVert=1$ and the cross-product of two non-null orthogonal vectors is never $0$.
Best Answer
Using circular shift property of cross product we get: $$\langle T \times B,N \rangle=\langle N \times T,B \rangle=-\langle T \times N,B \rangle=-1$$