Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t=0$:
${\bf{N}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 6 }},\sqrt {\frac{2}{3}} , - \frac{1}{{\sqrt 6 }}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t,{{\rm{e}}^t},t} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {1,{{\rm{e}}^t},1} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {0,{{\rm{e}}^t},0} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {2 + {{\rm{e}}^{2t}}} $
$v'\left( t \right) = \frac{{{{\rm{e}}^{2t}}}}{{\sqrt {2 + {{\rm{e}}^{2t}}} }}$
Evaluate the term
$v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t=0$:
$v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right) = \sqrt 3 \left( {0,1,0} \right) - \frac{1}{{\sqrt 3 }}\left( {1,1,1} \right)$
$v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right) = \left( { - \frac{1}{{\sqrt 3 }},\sqrt 3 - \frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t=0$ is given by
${\bf{N}}\left( 0 \right) = \frac{{v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right)}}{{||v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right)||}}$
${\bf{N}}\left( 0 \right) = \frac{{\left( { - \frac{1}{{\sqrt 3 }},\sqrt 3 - \frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)}}{{||\left( { - \frac{1}{{\sqrt 3 }},\sqrt 3 - \frac{1}{{\sqrt 3 }}, - \frac{1}{{\sqrt 3 }}} \right)||}}$
${\bf{N}}\left( 0 \right) = \left( { - \frac{1}{{\sqrt 6 }},\sqrt {\frac{2}{3}} , - \frac{1}{{\sqrt 6 }}} \right)$
