Matrix representation with respect to the dual basis. Let $T$ be a linear transformation, defined by:
$$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \\  T(x,y,z) = (x+y+z,\,y+z,\,z).$$
Let B = $\{v_1,v_2,v_3\}$ be a basis of $\mathbb{R}^3$. Given by the vectors,
$$v_1 = (1,1,1), v_2=(0,1,1), v_3=(0,0,1).$$
Calculate the matrix representation of the transponse $T^\top$ with respect to the dual basis 
$B^*$.
 A: Firstly, we need to find the dual basis $B^*$, which is the basis dual to $B$.
Since we want $v'_i(v_j) = \delta_{ij}$, notice that 
$$v'_1 \in \ker \operatorname{Span}(v_2,v_3)\;,\quad v'_2 \in \ker \operatorname{Span}(v_1,v_3)\;, \quad v'_3 \in \ker \operatorname{Span}(v_1,v_2); $$
in which $v_i (i=1,2,3)$ are row vectors and $v'_1,v'_2,v'_3$ are the normalized vectors of the kernels scaled such that $v'_i \cdot v_i =1$. So this gives our basis $B^* = \{v'_1,v'_2,v'_3\}$.
Next we want to construct $[T^\top]^{B^*}_{B^*}$. Now note 
$$[T^\top]^{B^*}_{B^*} = [\operatorname{id}]^E_{B^*} \cdot [T^\top]^E_E \cdot [\operatorname{id}]^{B^*}_E,$$
where $E$ is the standard basis and $[\operatorname{id}]^{B^*}_E$ is the basis transformation matrix from $B^*$ to $E$. In other words, $[\operatorname{id}]^{B^*}_E$ has $v'_1,v'_2,v'_3$ as columns. $[T^\top]^E_E$ represents the matrix representation of our (transposed) linear map with respect to the standard basis, so we have $$[T^\top]^E_E = \begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}.$$
Furthermore, we can compute $[\operatorname{id}]^E_{B^*}$, by using 
$$[\operatorname{id}]^E_{B^*} = \Big([\operatorname{id}]^{B^*}_E\Big)^{-1}.$$
We can now compute our desired $[T^\top]^{B^*}_{B^*}$.
Edit. Answer requested by the OP:
\begin{align*}
[T^\top]^{B^*}_{B^*} =& [\operatorname{id}]^E_{B^*} \cdot [T^\top]^E_E \cdot [\operatorname{id}]^{B^*}_E \\
=& \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix} \cdot
\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix} \cdot 
\begin{pmatrix}1&-1&0\\0&1&-1\\0&0&1\end{pmatrix} \\
=&^\text{Long live Wolfram!}\begin{pmatrix}3&-1&-1\\2&0&-1\\1&0&0\end{pmatrix}.
\end{align*}
