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

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(C) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$,and $\vec{c}=\hat{i}-\hat{j}$.First,we calculate the cro

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$(\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$

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(C) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$,and $\vec{c}=\hat{i}-\hat{j}$.First,we calculate the cross products:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 3 \end{vmatrix} = \hat{i}(3+1) - \hat{j}(3-2) + \hat{k}(-1-2) = 4\hat{i} - \hat{j} - 3\hat{k}$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 1 & -1 & 0 \end{vmatrix} = \hat{i}(0+3) - \hat{j}(0-3) + \hat{k}(-2+1) = 3\hat{i} + 3\hat{j} - \hat{k}$.$\vec{c} 	imes \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1+1) = -\hat{i} - \hat{j} + 2\hat{k}$.Given $6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(4\hat{i} - \hat{j} - 3\hat{k}) + \lambda_2(3\hat{i} + 3\hat{j} - \hat{k}) + \lambda_3(-\hat{i} - \hat{j} + 2\hat{k})$.Equating components,we get the system:$4\lambda_1 + 3\lambda_2 - \lambda_3 = 6$ $(1)$$-\lambda_1 + 3\lambda_2 - \lambda_3 = 2$ $(2)$$-3\lambda_1 - \lambda_2 + 2\lambda_3 = 3$ $(3)$Subtracting $(2)$ from $(1)$: $5\lambda_1 = 4 \implies \lambda_1 = \frac{4}{5}$.Substituting $\lambda_1$ into $(1)$ and $(3)$: $3\lambda_2 - \lambda_3 = 6 - \frac{16}{5} = \frac{14}{5}$ and $-\lambda_2 + 2\lambda_3 = 3 + \frac{12}{5} = \frac{27}{5}$.Solving these: $6\lambda_2 - 2\lambda_3 = \frac{28}{5}$ and $-\lambda_2 + 2\lambda_3 = \frac{27}{5}$.Adding: $5\lambda_2 = \frac{55}{5} = 11 \implies \lambda_2 = \frac{11}{5}$.Then $\lambda_3 = 3(\frac{11}{5}) - \frac{14}{5} = \frac{33-14}{5} = \frac{19}{5}$.Thus,$(\lambda_1, \lambda_2, \lambda_3) = (\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$

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