Answer
$${\bf{N}}\left( \pi \right) = {\bf{i}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \cos 3t{\bf{i}} + 2\sin 3t{\bf{j}} + {\bf{k}},{\text{ }}t = \pi \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = - 3\sin 3t{\bf{i}} + 6\cos 3t{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{ - 3\sin 3t{\bf{i}} + 6\cos 3t{\bf{j}}}}{{\left\| { - 3\sin 3t{\bf{i}} + 6\cos 3t{\bf{j}}} \right\|}} = \frac{{ - 3\sin 3t{\bf{i}} + 6\cos 3t{\bf{j}}}}{{\sqrt {9{{\sin }^2}3t + 36{{\cos }^2}3t} }} \cr
& {\bf{T}}\left( t \right) = \frac{{ - 3\sin 3t{\bf{i}} + 6\cos 3t{\bf{j}}}}{{\sqrt {9\left( {{{\sin }^2}3t + 4{{\cos }^2}3t} \right)} }} \cr
& {\bf{T}}\left( t \right) = \frac{{ - \sin 3t{\bf{i}} + 2\cos 3t{\bf{j}}}}{{3\sqrt {{{\sin }^2}3t + 4{{\cos }^2}3t} }} \cr
& {\bf{T}}\left( t \right) = - \frac{{\sin 3t}}{{3\sqrt {{{\sin }^2}3t + 4{{\cos }^2}3t} }}{\bf{i}} + \frac{{2\cos 3t}}{{3\sqrt {{{\sin }^2}3t + 4{{\cos }^2}3t} }}{\bf{j}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\text{Making the operations by hand and replacing }}\left( {{\text{or use a CAS}}} \right) \cr
& {\bf{T}}'\left( t \right) = - \frac{{4\cos 3t}}{{{{\left( {{{\sin }^2}3t + 4{{\cos }^2}3t} \right)}^{3/2}}}}{\bf{i}} - \frac{{2\sin 3t}}{{{{\left( {{{\sin }^2}3t + 4{{\cos }^2}3t} \right)}^{3/2}}}}{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = \pi \cr
& {\bf{T}}'\left( \pi \right) = - \frac{{4\cos 3\pi }}{{{{\left( {{{\sin }^2}3\pi + 4{{\cos }^2}3\pi } \right)}^{3/2}}}}{\bf{i}} - \frac{{2\sin 3\pi }}{{{{\left( {{{\sin }^2}3\pi + 4{{\cos }^2}3\pi } \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( \pi \right) = - \frac{{ - 4}}{{{{\left( 4 \right)}^{3/2}}}}{\bf{i}} - \frac{0}{{{{\left( 4 \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( \pi \right) = \frac{1}{{\sqrt 4 }}{\bf{i}} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = \pi \cr
& {\bf{N}}\left( \pi \right) = \frac{{{\bf{T}}'\left( \pi \right)}}{{\left\| {{\bf{T}}'\left( \pi \right)} \right\|}} = \frac{{\frac{1}{{\sqrt 4 }}{\bf{i}}}}{{\left\| {\frac{1}{{\sqrt 4 }}{\bf{i}}} \right\|}} \cr
& {\bf{N}}\left( \pi \right) = {\bf{i}} \cr} $$
