Let $\overrightarrow{\mathrm{a}}=\hat{i}+2 \hat{j}+\hat{k}, $
$\overrightarrow{\mathrm{b}}=3(\hat{i}-\hat{j}+\hat{k})$.
Let $\overrightarrow{\mathrm{c}}$ be the vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ and $\vec{a} \cdot \vec{c}=3$.
Then $\vec{a} \cdot((\vec{c} \times \vec{b})-\vec{b}-\vec{c})$ is equal to :

<p>$$\begin{aligned} & \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\ & \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { given } \vec{a} \times \vec{c}=\vec{b} \\ & \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\ & \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\ & \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)} \end{aligned}$$</p> <p>$$\begin{gathered} \text { By (i), (ii), (iii) & (iv) } \\ 27-0-3=24 \end{gathered}$$</p>