Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $\vec{a} \cdot \vec{c}=-29$, then $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

<p>$$\begin{gathered} (\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\ \Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\ =\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\ \quad=\lambda(7 \hat{i}+8 \hat{j}) \\ \Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\ \quad \vec{c} \cdot \vec{a}=-29 \\ \Rightarrow \quad 14 \lambda-2+40 \lambda=-29 \\ \Rightarrow \quad 54 \lambda=-27 \\ \Rightarrow \quad \lambda=-\frac{1}{2} \end{gathered}$$</p> <p>$$\begin{aligned} & \therefore \quad \vec{c}=\left(\frac{-7}{2}-1\right) \hat{i}-4 \hat{j}=\frac{-9}{2} \hat{i}-4 \hat{j} \\ & \vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=9-4=5 \end{aligned}$$</p>