Let the vectors vec{a} = -hat{i} + hat{j} + 3 | vedclass

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(B) Given $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$.$\vec{c} = \lambda(-\hat{i} + \hat{j} + 3\hat{k}) + \

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$12$

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(B) Given $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$.$\vec{c} = \lambda(-\hat{i} + \hat{j} + 3\hat{k}) + \mu(\hat{i} + 3\hat{j} + \hat{k}) = (\mu-\lambda)\hat{i} + (\lambda+3\mu)\hat{j} + (3\lambda+\mu)\hat{k}$.Given $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$, we have $3(\mu-\lambda) - 6(\lambda+3\mu) + 2(3\lambda+\mu) = 10$.$3\mu - 3\lambda - 6\lambda - 18\mu + 6\lambda + 2\mu = 10 \Rightarrow -3\lambda - 13\mu = 10$  (Equation $1$).Given $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, we have $(\mu-\lambda) + (\lambda+3\mu) + (3\lambda+\mu) = -2$.$3\lambda + 5\mu = -2$  (Equation $2$).Adding Equation $1$ and Equation $2$: $(-3\lambda - 13\mu) + (3\lambda + 5\mu) = 10 - 2 \Rightarrow -8\mu = 8 \Rightarrow \mu = -1$.Substituting $\mu = -1$ into Equation $2$: $3\lambda + 5(-1) = -2 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$.Thus, $\vec{c} = 1(-\hat{i} + \hat{j} + 3\hat{k}) - 1(\hat{i} + 3\hat{j} + \hat{k}) = -2\hat{i} - 2\hat{j} + 2\hat{k}$.$|\vec{c}|^2 = (-2)^2 + (-2)^2 + 2^2 = 4 + 4 + 4 = 12$.

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