Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and a vector $\vec{c}$ be such that $$\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$$. If $\vec{a} \cdot \vec{c}=13$, then $(24-\vec{b} \cdot \vec{c})$ is equal to _______.

<p>Let $\hat{i}+8 \hat{j}+13 \hat{k}=\vec{u}$</p> <p>Given $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\vec{u}$</p> <p>$$\begin{gathered} \Rightarrow \quad \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\vec{u} \\ (\vec{a}+\vec{b}) \times c=\vec{u}-\vec{a} \times \vec{b} \end{gathered}$$</p> <p>Taking cross product with $\vec{a}$ on both sides</p> <p>$$\vec{a} \times((\vec{a}+\vec{b}) \times \vec{c})=\vec{a} \times(\vec{u}-\vec{a} \times \vec{b})$$</p> <p>$$\Rightarrow \quad \vec{c} \cdot\left(\vec{a}^2+\vec{a} \cdot \vec{b}\right)=13(\vec{a}+\vec{b})-\vec{a} \times \vec{u} +(\vec{a} \cdot \vec{b}) \cdot \vec{a}-\vec{a}^2 \vec{b}\quad$$ $\{\because \vec{a} . \vec{c}=13\}$</p> <p>Putting the values, $\vec{c}=(-1,-1,3)$</p> <p>$\vec{b}.\vec{c}=-22$</p> <p>$\Rightarrow 24-\vec{b} \cdot \vec{c}=46$</p>