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(C) Given $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$.Area of the triangle $= \frac{1}{2} |\vec{a} 	i

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$2$

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(C) Given $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$.Area of the triangle $= \frac{1}{2} |\vec{a} 	imes \vec{b}| = \frac{\sqrt{195}}{2}$.Thus,$|\vec{a} 	imes \vec{b}|^2 = 195$ ...$(i)$.Now,$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & \lambda \ \lambda & 1 & -3 \end{vmatrix} = \hat{i}(3-\lambda) - \hat{j}(-9-\lambda^2) + \hat{k}(3+\lambda) = (3-\lambda)\hat{i} + (9+\lambda^2)\hat{j} + (3+\lambda)\hat{k}$.$|\vec{a} 	imes \vec{b}|^2 = (3-\lambda)^2 + (9+\lambda^2)^2 + (3+\lambda)^2 = 195$.Expanding the terms: $(9 + \lambda^2 - 6\lambda) + (81 + \lambda^4 + 18\lambda^2) + (9 + \lambda^2 + 6\lambda) = 195$.$\lambda^4 + 20\lambda^2 + 99 = 195 \Rightarrow \lambda^4 + 20\lambda^2 - 96 = 0$.Let $t = \lambda^2$,then $t^2 + 20t - 96 = 0 \Rightarrow (t+24)(t-4) = 0$.Since $\lambda$ is a real number,$\lambda^2 = 4$ (as $\lambda^2 = -24$ is impossible).Thus,$\lambda = \pm 2$. The number of distinct values of $\lambda$ is $2$.

For some real number $\lambda$,if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$,then the number of distinct possible values of $\lambda$ is

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