Let $$\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$$

Given $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$$

$$\begin{gathered} \Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\ (\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b} \end{gathered}$$

Taking cross product with $$\vec{a}$$ on both sides

$$\vec{a} \times((\vec{a}+\vec{b}) \times \vec{c})=\vec{a} \times(\vec{u}-\vec{a} \times \vec{b})$$

$$\Rightarrow \quad \vec{c} \cdot\left(\vec{a}^2+\vec{a} \cdot \vec{b}\right)=13(\vec{a}+\vec{b})-\vec{a} \times \vec{u} +(\vec{a} \cdot \vec{b}) \cdot \vec{a}-\vec{a}^2 \vec{b}\quad$$ $$\{\because \vec{a} . \vec{c}=13\}$$

Putting the values, $$\vec{c}=(-1,-1,3)$$

$$\vec{b}.\vec{c}=-22$$

$$\Rightarrow 24-\vec{b} \cdot \vec{c}=46$$