Answer
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 1 \right) = \left( {0,\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)}\\
{{\bf{N}}\left( 1 \right) = \left( {0, - \frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)}\\
{{\bf{B}}\left( 1 \right) = \left( {1,0,0} \right)}
\end{array}$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {0,t,{t^2}} \right)$. The point $\left( {0,1,1} \right)$ corresponds to $t=1$.
The tangent vector is ${\bf{r}}'\left( t \right) = \left( {0,1,2t} \right)$. So, the unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {0,1,2t} \right)}}{{\sqrt {\left( {0,1,2t} \right)\cdot\left( {0,1,2t} \right)} }}$
${\bf{T}}\left( t \right) = \left( {0,\frac{1}{{\sqrt {1 + 4{t^2}} }},\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}} \right)$
At $t=1$, we get ${\bf{T}}\left( 1 \right) = \left( {0,\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)$.
To evaluate ${\bf{T}}'\left( t \right)$:
1. Evaluate $\frac{d}{{dt}}\frac{1}{{\sqrt {1 + 4{t^2}} }}$
$\frac{d}{{dt}}\frac{1}{{\sqrt {1 + 4{t^2}} }} = \frac{d}{{dt}}\left( {{{\left( {1 + 4{t^2}} \right)}^{ - 1/2}}} \right)$
$\frac{d}{{dt}}\frac{1}{{\sqrt {1 + 4{t^2}} }} = - \frac{1}{2}\left( {8t} \right){\left( {1 + 4{t^2}} \right)^{ - 3/2}} = - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}$
2. Evaluate $\frac{d}{{dt}}\frac{{2t}}{{\sqrt {1 + 4{t^2}} }}$
$\frac{d}{{dt}}\frac{{2t}}{{\sqrt {1 + 4{t^2}} }} = \frac{{2\sqrt {1 + 4{t^2}} - \left( {2t} \right)\left( {\frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{1/2}}}}} \right)}}{{1 + 4{t^2}}}$
$\frac{d}{{dt}}\frac{{2t}}{{\sqrt {1 + 4{t^2}} }} = \frac{{2\left( {1 + 4{t^2}} \right) - 8{t^2}}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}} = \frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}$
Thus, ${\bf{T}}'\left( t \right) = \left( {0, - \frac{{4t}}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}},\frac{2}{{{{\left( {1 + 4{t^2}} \right)}^{3/2}}}}} \right)$
At $t=1$, we get
${\bf{T}}'\left( 1 \right) = \left( {0, - \frac{4}{{{5^{3/2}}}},\frac{2}{{{5^{3/2}}}}} \right)$
$||{\bf{T}}'\left( 1 \right)|{|^2} = \frac{{16}}{{{5^3}}} + \frac{4}{{{5^3}}} = \frac{{20}}{{{5^3}}}$
$||{\bf{T}}'\left( 1 \right)|| = \frac{{\sqrt {20} }}{{{5^{3/2}}}}$
The normal vector at $t=1$,
${\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{||{\bf{T}}'\left( 1 \right)||}}$
${\bf{N}}\left( 1 \right) = \frac{{{5^{3/2}}}}{{\sqrt {20} }}\left( {0, - \frac{4}{{{5^{3/2}}}},\frac{2}{{{5^{3/2}}}}} \right) = \left( {0, - \frac{4}{{\sqrt {20} }},\frac{2}{{\sqrt {20} }}} \right)$
${\bf{N}}\left( 1 \right) = \left( {0, - \frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)$
By Eq. (8), the binormal vector at $t=1$ is given by
${\bf{B}}\left( 1 \right) = {\bf{T}}\left( 1 \right) \times {\bf{N}}\left( 1 \right)$
${\bf{B}}\left( 1 \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&{\frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}}\\
0&{ - \frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}
\end{array}} \right|$
${\bf{B}}\left( 1 \right) = \left( {\frac{1}{5} + \frac{4}{5}} \right){\bf{i}} = {\bf{i}}$, ${\ \ \ }$ ${\bf{B}}\left( 1 \right) = \left( {1,0,0} \right)$
In summary, we obtain
$\begin{array}{*{20}{c}}
{{\bf{T}}\left( 1 \right) = \left( {0,\frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)}\\
{{\bf{N}}\left( 1 \right) = \left( {0, - \frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)}\\
{{\bf{B}}\left( 1 \right) = \left( {1,0,0} \right)}
\end{array}$
