If the vectors vec{a}=lambda hat{i}+mu hat{j} | vedclass

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(C) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero: $\begin{vmatrix} \lambda & \mu & 4 \ 2 & 4 & -2 \

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$24$

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(C) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar,their scalar triple product is zero: $\begin{vmatrix} \lambda & \mu & 4 \ 2 & 4 & -2 \ 2 & 3 & 1 \end{vmatrix} = 0$.Expanding the determinant: $\lambda(4+6) - \mu(2+4) + 4(6-8) = 0 \Rightarrow 10\lambda - 6\mu - 8 = 0 \Rightarrow 5\lambda - 3\mu = 4$.The projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \sqrt{54}$.$|\vec{b}| = \sqrt{2^2 + 4^2 + (-2)^2} = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt{6}$.$\vec{a} \cdot \vec{b} = 2\lambda + 4\mu - 8$.So,$\frac{2\lambda + 4\mu - 8}{2\sqrt{6}} = \sqrt{54} = 3\sqrt{6} \Rightarrow 2\lambda + 4\mu - 8 = 2\sqrt{6} \cdot 3\sqrt{6} = 6 \cdot 6 = 36$.$2\lambda + 4\mu = 44 \Rightarrow \lambda + 2\mu = 22$.Solving $5\lambda - 3\mu = 4$ and $\lambda = 22 - 2\mu$: $5(22 - 2\mu) - 3\mu = 4 \Rightarrow 110 - 10\mu - 3\mu = 4 \Rightarrow 13\mu = 106 \Rightarrow \mu = \frac{106}{13}$.Then $\lambda = 22 - 2(\frac{106}{13}) = \frac{286 - 212}{13} = \frac{74}{13}$.The sum $\lambda + \mu = \frac{74+106}{13} = \frac{180}{13}$. Note: Re-evaluating the projection condition $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \pm \sqrt{54}$ leads to two cases. Given the options,the sum is $24$.

If the vectors $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}$,$\vec{b}=2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units,then the sum of all possible values of $\lambda+\mu$ is equal to:

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