The sum of all values of alpha,for which the | vedclass

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(D) Let the given points be $P, Q, R,$ and $S$ with position vectors:$\vec{p} = \hat{i}-2 \hat{j}+3 \hat{k}$$\vec{q} = 2 \hat{i}-3 \hat{j}+4 \hat{k}$$

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$2$

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(D) Let the given points be $P, Q, R,$ and $S$ with position vectors:$\vec{p} = \hat{i}-2 \hat{j}+3 \hat{k}$$\vec{q} = 2 \hat{i}-3 \hat{j}+4 \hat{k}$$\vec{r} = (\alpha+1) \hat{i}+2 \hat{k}$$\vec{s} = 9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$The points are coplanar if the vectors $\vec{PQ}, \vec{PR},$ and $\vec{PS}$ are coplanar,which means their scalar triple product is zero:$\vec{PQ} = \vec{q} - \vec{p} = (2-1)\hat{i} + (-3 - (-2))\hat{j} + (4-3)\hat{k} = \hat{i} - \hat{j} + \hat{k}$$\vec{PR} = \vec{r} - \vec{p} = (\alpha+1-1)\hat{i} + (0 - (-2))\hat{j} + (2-3)\hat{k} = \alpha\hat{i} + 2\hat{j} - \hat{k}$$\vec{PS} = \vec{s} - \vec{p} = (9-1)\hat{i} + (\alpha-8 - (-2))\hat{j} + (6-3)\hat{k} = 8\hat{i} + (\alpha-6)\hat{j} + 3\hat{k}$For coplanarity,the determinant of these vectors must be zero:$\begin{vmatrix} 1 & -1 & 1 \ \alpha & 2 & -1 \ 8 & \alpha-6 & 3 \end{vmatrix} = 0$Expanding the determinant:$1(6 - (-1)(\alpha-6)) - (-1)(3\alpha - (-8)) + 1(\alpha(\alpha-6) - 16) = 0$$1(6 + \alpha - 6) + 1(3\alpha + 8) + (\alpha^2 - 6\alpha - 16) = 0$$\alpha + 3\alpha + 8 + \alpha^2 - 6\alpha - 16 = 0$$\alpha^2 - 2\alpha - 8 = 0$$(\alpha - 4)(\alpha + 2) = 0$Thus,the values of $\alpha$ are $4$ and $-2$.The sum of these values is $4 + (-2) = 2$.

The sum of all values of $\alpha$,for which the points whose position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$,$2 \hat{i}-3 \hat{j}+4 \hat{k}$,$(\alpha+1) \hat{i}+2 \hat{k}$ and $9 \hat{i}+(\alpha-8) \hat{j}+6 \hat{k}$ are coplanar,is equal to

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