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(A) The points $A, B, C, D$ are coplanar if and only if the scalar triple product of the vectors $\vec{AB}, \vec{AC}, \vec{AD}$ is zero,i.e.,$[\vec{AB

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(A) The points $A, B, C, D$ are coplanar if and only if the scalar triple product of the vectors $\vec{AB}, \vec{AC}, \vec{AD}$ is zero,i.e.,$[\vec{AB}, \vec{AC}, \vec{AD}] = 0$.First,we find the vectors:$\vec{AB} = (1-5)\hat{i} + (2-5)\hat{j} + (3-2\lambda)\hat{k} = -4\hat{i} - 3\hat{j} + (3-2\lambda)\hat{k}$$\vec{AC} = (-2-5)\hat{i} + (\lambda-5)\hat{j} + (4-2\lambda)\hat{k} = -7\hat{i} + (\lambda-5)\hat{j} + (4-2\lambda)\hat{k}$$\vec{AD} = (-1-5)\hat{i} + (5-5)\hat{j} + (6-2\lambda)\hat{k} = -6\hat{i} + 0\hat{j} + (6-2\lambda)\hat{k}$Setting the determinant to zero:$\begin{vmatrix} -4 & -3 & 3-2\lambda \ -7 & \lambda-5 & 4-2\lambda \ -6 & 0 & 6-2\lambda \end{vmatrix} = 0$Expanding along the third row:$-6[(-3)(4-2\lambda) - (3-2\lambda)(\lambda-5)] + (6-2\lambda)[(-4)(\lambda-5) - (-3)(-7)] = 0$$-6[-12 + 6\lambda - (3\lambda - 15 - 2\lambda^2 + 10\lambda)] + (6-2\lambda)[-4\lambda + 20 - 21] = 0$$-6[2\lambda^2 - 7\lambda + 3] + (6-2\lambda)(-4\lambda - 1) = 0$$-12\lambda^2 + 42\lambda - 18 - 24\lambda - 6 + 8\lambda^2 + 2\lambda = 0$$-4\lambda^2 + 20\lambda - 24 = 0$$\lambda^2 - 5\lambda + 6 = 0$$(\lambda-2)(\lambda-3) = 0$So,$S = \{2, 3\}$.Calculating the sum: $\sum_{\lambda \in S}(\lambda+2)^2 = (2+2)^2 + (3+2)^2 = 4^2 + 5^2 = 16 + 25 = 41$.

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