If vec{a}=hat{i}+2 hat{j}+hat{k}, vec{b}=3(ha | vedclass

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(B) Given $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$.Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.From $\vec{a}

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$24$

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(B) Given $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}-3 \hat{j}+3 \hat{k}$.Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.From $\vec{a} 	imes \vec{c}=\vec{b}$, we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ x & y & z \end{vmatrix} = 3 \hat{i}-3 \hat{j}+3 \hat{k}$$(2z-y) \hat{i} - (z-x) \hat{j} + (y-2x) \hat{k} = 3 \hat{i}-3 \hat{j}+3 \hat{k}$Comparing components:$2z-y=3$ $(i)$$x-z=-3 \Rightarrow z-x=3$ (ii)$y-2x=3$ (iii)Given $\vec{a} \cdot \vec{c}=3$, so $x+2y+z=3$ (iv).From (ii), $z=x+3$. Substituting into (iv): $x+2y+x+3=3 \Rightarrow 2x+2y=0 \Rightarrow y=-x$.Substituting $y=-x$ into (iii): $-x-2x=3 \Rightarrow -3x=3 \Rightarrow x=-1$.Then $y=1$ and $z=2$. Thus, $\vec{c}=-\hat{i}+\hat{j}+2 \hat{k}$.Now, $\vec{c} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 2 \ 3 & -3 & 3 \end{vmatrix} = (3+6) \hat{i} - (-3-6) \hat{j} + (3-3) \hat{k} = 9 \hat{i}+9 \hat{j}$.We need to calculate $\vec{a} \cdot (\vec{c} 	imes \vec{b} - \vec{b} - \vec{c}) = \vec{a} \cdot (\vec{c} 	imes \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$.$\vec{a} \cdot (\vec{c} 	imes \vec{b}) = (\hat{i}+2 \hat{j}+\hat{k}) \cdot (9 \hat{i}+9 \hat{j}) = 9+18=27$.$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3-6+3=0$.$\vec{a} \cdot \vec{c} = 3$.Therefore, $27 - 0 - 3 = 24$.

If $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$, then $\vec{a} \cdot(\vec{c} \times \vec{b}-\vec{b}-\vec{c})=$

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