已知向量 $\alpha_{1}=\left(\begin{array}{l}1 \\ 0 \\ 1 \\ 1\end{array}\right), \alpha_{2}=\left(\begin{array}{l}-1 \\ -1 \\ 0 \\ 1\end{array}\right), \alpha_{3}=\left(\begin{array}{l}0 \\ 1 \\ -1 \\ 1\end{array}\right), \beta=\left(\begin{array}{l}1 \\ 1 \\ 1 \\ -1\end{array}\right), \gamma=k_{1} \alpha_{1}+k_{2} \alpha_{2}+k_{3} \alpha_{3}$,
若 $\gamma^{T} \alpha_{i}=\beta^{T} \alpha_{i}(i=1,2,3)$, 则 $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=$
$\gamma^{T} \alpha_{1}=\beta^{T} \alpha_{1}=1 \Rightarrow k_{1} \alpha_{1}^{T} \alpha_{1}+k_{2} \alpha_{2}^{T} \alpha_{1}+k_{3} \alpha_{3}^{T} \alpha_{1}=1 \Rightarrow k_{1} \cdot 3+k_{2} \cdot 0+k_{3} \cdot 0=1 \Rightarrow k_{1}=\frac{1}{3}$.
同理 $k_{2}=-1, k_{3}=-\frac{1}{3}$.
所以, $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=\frac{11}{9}$.

因此答案为：$\frac{11}{9}$