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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$.We are given $\vec{a} 	imes \vec{c} = \vec{b}$ and $\vec{a} \cdot

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$-2$

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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$.We are given $\vec{a} 	imes \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$.We need to find $\vec{a} \cdot (\vec{b} 	imes \vec{c})$.Using the scalar triple product property,$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = (\vec{a} 	imes \vec{b}) \cdot \vec{c}$.First,calculate $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1-1) - \hat{j}(-1-0) + \hat{k}(1-0) = -2\hat{i} + \hat{j} + \hat{k}$.Now,use the vector triple product identity $\vec{a} 	imes (\vec{a} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.Since $\vec{a} 	imes \vec{c} = \vec{b}$,we have $\vec{a} 	imes \vec{b} = (\vec{a} \cdot \vec{c})\vec{a} - |\vec{a}|^2 \vec{c}$.Given $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$ and $\vec{a} \cdot \vec{c} = 3$,we get:$-2\hat{i} + \hat{j} + \hat{k} = 3(\hat{i} + \hat{j} + \hat{k}) - 3\vec{c}$.$3\vec{c} = 3\hat{i} + 3\hat{j} + 3\hat{k} - (-2\hat{i} + \hat{j} + \hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$.$\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.Finally,$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = (\vec{a} 	imes \vec{b}) \cdot \vec{c} = (-2\hat{i} + \hat{j} + \hat{k}) \cdot (\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}) = -\frac{10}{3} + \frac{2}{3} + \frac{2}{3} = -\frac{6}{3} = -2$.

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

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