If the four points 2vec{a} + 3vec{b} - vec{c} | vedclass

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(D) Let the four points be $A, B, C,$ and $D$ represented by position vectors $\vec{p_1} = 2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{p_2} = \vec{a} - 2\vec

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(D) Let the four points be $A, B, C,$ and $D$ represented by position vectors $\vec{p_1} = 2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{p_2} = \vec{a} - 2\vec{b} + 3\vec{c}$,$\vec{p_3} = 3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{p_4} = \vec{a} - \lambda\vec{b} - 6\vec{c}$.For the points to be coplanar,the vectors $\vec{AB}, \vec{AC},$ and $\vec{AD}$ must be coplanar,meaning their scalar triple product is zero: $[\vec{AB}, \vec{AC}, \vec{AD}] = 0$.$\vec{AB} = \vec{p_2} - \vec{p_1} = -\vec{a} - 5\vec{b} + 4\vec{c}$$\vec{AC} = \vec{p_3} - \vec{p_1} = \vec{a} + \vec{b} - \vec{c}$$\vec{AD} = \vec{p_4} - \vec{p_1} = -\vec{a} - (\lambda + 3)\vec{b} - 5\vec{c}$The scalar triple product is given by the determinant:$\begin{vmatrix} -1 & -5 & 4 \ 1 & 1 & -1 \ -1 & -(\lambda + 3) & -5 \end{vmatrix} = 0$Expanding the determinant:$-1(-5 - (\lambda + 3)) + 5(-5 - 1) + 4(-(\lambda + 3) + 1) = 0$$-1(-5 - \lambda - 3) + 5(-6) + 4(-\lambda - 3 + 1) = 0$$-1(-\lambda - 8) - 30 + 4(-\lambda - 2) = 0$$\lambda + 8 - 30 - 4\lambda - 8 = 0$$-3\lambda - 30 = 0$$-3\lambda = 30 \implies \lambda = -10$.Since $-10$ is not among the options,the correct answer is $D$.

If the four points $2\vec{a} + 3\vec{b} - \vec{c}$,$\vec{a} - 2\vec{b} + 3\vec{c}$,$3\vec{a} + 4\vec{b} - 2\vec{c}$,and $\vec{a} - \lambda\vec{b} - 6\vec{c}$ are coplanar,find the value of $\lambda$.

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