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

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(A) Given the vector equation $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$.Substituting the given vectors,we have $\alpha

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$(2, -1, -1)$

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(A) Given the vector equation $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$.Substituting the given vectors,we have $\alpha(1, 2, 3) + \beta(2, 3, 1) + \gamma(3, 1, 2) = -3(1, 0, -1)$.This results in the system of linear equations:$1) \alpha + 2\beta + 3\gamma = -3$$2) 2\alpha + 3\beta + \gamma = 0$$3) 3\alpha + \beta + 2\gamma = 3$Adding all three equations: $(\alpha + 2\alpha + 3\alpha) + (2\beta + 3\beta + \beta) + (3\gamma + \gamma + 2\gamma) = -3 + 0 + 3 \Rightarrow 6\alpha + 6\beta + 6\gamma = 0 \Rightarrow \alpha + \beta + \gamma = 0$.From equation $(2)$,$\gamma = -2\alpha - 3\beta$. Substituting this into $\alpha + \beta + \gamma = 0$ gives $\alpha + \beta - 2\alpha - 3\beta = 0 \Rightarrow -\alpha - 2\beta = 0 \Rightarrow \alpha = -2\beta$.Substituting $\alpha = -2\beta$ into equation $(1)$: $(-2\beta) + 2\beta + 3\gamma = -3 \Rightarrow 3\gamma = -3 \Rightarrow \gamma = -1$.Since $\alpha + \beta + \gamma = 0$,we have $-2\beta + \beta - 1 = 0 \Rightarrow -\beta = 1 \Rightarrow \beta = -1$.Then $\alpha = -2(-1) = 2$.Thus,the triplet $(\alpha, \beta, \gamma)$ is $(2, -1, -1)$.

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = -3(\hat{i} - \hat{k})$. Then the triplet $(\alpha, \beta, \gamma)$ is

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