Let vec{a}, vec{b} and vec{c} be three non-ze | vedclass

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(C) The position vectors are given as:$\vec{OA} = \vec{a} - \vec{b} + \vec{c}$$\vec{OB} = \lambda \vec{a} - 3 \vec{b} + 4 \vec{c}$$\vec{OC} = -\vec{a}

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$2$

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(C) The position vectors are given as:$\vec{OA} = \vec{a} - \vec{b} + \vec{c}$$\vec{OB} = \lambda \vec{a} - 3 \vec{b} + 4 \vec{c}$$\vec{OC} = -\vec{a} + 2 \vec{b} - 3 \vec{c}$$\vec{OD} = 2 \vec{a} - 4 \vec{b} + 6 \vec{c}$Now,calculate the vectors $\overrightarrow{AB}$,$\overrightarrow{AC}$,and $\overrightarrow{AD}$:$\overrightarrow{AB} = \vec{OB} - \vec{OA} = (\lambda - 1)\vec{a} - 2\vec{b} + 3\vec{c}$$\overrightarrow{AC} = \vec{OC} - \vec{OA} = -2\vec{a} + 3\vec{b} - 4\vec{c}$$\overrightarrow{AD} = \vec{OD} - \vec{OA} = \vec{a} - 3\vec{b} + 5\vec{c}$Since $\overrightarrow{AB}$,$\overrightarrow{AC}$,and $\overrightarrow{AD}$ are coplanar,their scalar triple product must be zero:$\begin{vmatrix} \lambda - 1 & -2 & 3 \ -2 & 3 & -4 \ 1 & -3 & 5 \end{vmatrix} = 0$Expanding the determinant along the first row:$(\lambda - 1)(3 	imes 5 - (-4) 	imes (-3)) - (-2)((-2) 	imes 5 - (-4) 	imes 1) + 3((-2) 	imes (-3) - 3 	imes 1) = 0$$(\lambda - 1)(15 - 12) + 2(-10 + 4) + 3(6 - 3) = 0$$(\lambda - 1)(3) + 2(-6) + 3(3) = 0$$3\lambda - 3 - 12 + 9 = 0$$3\lambda - 6 = 0$$3\lambda = 6 \Rightarrow \lambda = 2$

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