If vec{a}, vec{b}, vec{c} are non-coplanar ve | vedclass

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(C) Let the position vectors of the four points be $\vec{p_1}, \vec{p_2}, \vec{p_3}, \vec{p_4}$. The points are coplanar if the vectors $\vec{p_2}-\ve

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$2$

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(C) Let the position vectors of the four points be $\vec{p_1}, \vec{p_2}, \vec{p_3}, \vec{p_4}$. The points are coplanar if the vectors $\vec{p_2}-\vec{p_1}$, $\vec{p_3}-\vec{p_1}$, and $\vec{p_4}-\vec{p_1}$ are coplanar. These vectors are: $\vec{v_1} = \vec{p_2}-\vec{p_1} = (1-\lambda)\vec{a} - (\lambda+3)\vec{b} + 4\vec{c}$ $\vec{v_2} = \vec{p_3}-\vec{p_1} = (3-\lambda)\vec{a} + 1\vec{b} + (1-\lambda)\vec{c}$ $\vec{v_3} = \vec{p_4}-\vec{p_1} = (1-\lambda)\vec{a} - 9\vec{b} + 7\vec{c}$ For these to be coplanar, the scalar triple product must be zero: $\begin{vmatrix} 1-\lambda & -(\lambda+3) & 4 \ 3-\lambda & 1 & 1-\lambda \ 1-\lambda & -9 & 7 \end{vmatrix} = 0$ Expanding the determinant, we find that $\lambda = 2$ satisfies the equation.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and the points $P_1 = \lambda \vec{a}+3 \vec{b}-\vec{c}$, $P_2 = \vec{a}-\lambda \vec{b}+3 \vec{c}$, $P_3 = 3 \vec{a}+4 \vec{b}-\lambda \vec{c}$, and $P_4 = \vec{a}-6 \vec{b}+6 \vec{c}$ are coplanar, then one of the values of $\lambda$ is

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