Key Concepts and Formulas

• Coplanarity of Four Points: Four points $A, B, C, D$ are coplanar if the vectors formed by taking one point as the origin and connecting it to the other three points are coplanar. For example, if we choose point $A$, then $\vec{AB}, \vec{AC}, \vec{AD}$ must be coplanar.
• Coplanarity of Three Vectors: Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product (STP) is zero: $[\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) = 0$.
• Scalar Triple Product as a Determinant: The scalar triple product of three vectors with components $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ is given by the determinant: $\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix}$
• Vector Subtraction: If $\vec{OA}$ and $\vec{OB}$ are position vectors of points $A$ and $B$, then $\vec{AB} = \vec{OB} - \vec{OA}$.

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Step-by-Step Solution

Step 1: Define the Position Vectors of the Given Points Let the four given points be $P_1, P_2, P_3, P_4$. Their position vectors from the origin $O$ are:

• $\vec{OP_1} = \hat{i} - 2\hat{j} + 3\hat{k}$
• $\vec{OP_2} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
• $\vec{OP_3} = (\alpha+1)\hat{i} + 0\hat{j} + 2\hat{k}$ (Note: the coefficient of $\hat{j}$ is 0)
• $\vec{OP_4} = 9\hat{i} + (\alpha-8)\hat{j} + 6\hat{k}$

Step 2: Form Three Vectors from a Common Point To check for coplanarity, we form three vectors originating from one of the points. Let's choose $P_1$ as the common point. We will compute the vectors $\vec{P_1P_2}$, $\vec{P_1P_3}$, and $\vec{P_1P_4}$.

• Vector $\vec{P_1P_2}$: $\vec{P_1P_2} = \vec{OP_2} - \vec{OP_1} = (2\hat{i} - 3\hat{j} + 4\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$ $\vec{P_1P_2} = (2-1)\hat{i} + (-3 - (-2))\hat{j} + (4-3)\hat{k} = \hat{i} - \hat{j} + \hat{k}$

• Vector $\vec{P_1P_3}$: $\vec{P_1P_3} = \vec{OP_3} - \vec{OP_1} = ((\alpha+1)\hat{i} + 2\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$ $\vec{P_1P_3} = ((\alpha+1)-1)\hat{i} - (0 - (-2))\hat{j} + (2-3)\hat{k} = \alpha\hat{i} - 2\hat{j} - \hat{k}$ Correction: Let's re-calculate $\vec{P_1P_3}$ carefully. $\vec{P_1P_3} = ((\alpha+1)\hat{i} + 0\hat{j} + 2\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$ $\vec{P_1P_3} = ((\alpha+1)-1)\hat{i} + (0 - (-2))\hat{j} + (2-3)\hat{k} = \alpha\hat{i} + 2\hat{j} - \hat{k}$ This matches the original solution's calculation for $\vec{AC}$ which corresponds to $\vec{P_1P_3}$.

• Vector $\vec{P_1P_4}$: $\vec{P_1P_4} = \vec{OP_4} - \vec{OP_1} = (9\hat{i} + (\alpha-8)\hat{j} + 6\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$ $\vec{P_1P_4} = (9-1)\hat{i} + ((\alpha-8) - (-2))\hat{j} + (6-3)\hat{k}$ $\vec{P_1P_4} = 8\hat{i} + (\alpha-8+2)\hat{j} + 3\hat{k} = 8\hat{i} + (\alpha-6)\hat{j} + 3\hat{k}$

Step 3: Apply the Coplanarity Condition Since the four points are coplanar, the three vectors $\vec{P_1P_2}$, $\vec{P_1P_3}$, and $\vec{P_1P_4}$ must be coplanar. Their scalar triple product must be zero. We express this using a determinant: $\begin{vmatrix} 1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha-6 & 3 \end{vmatrix} = 0$

Step 4: Evaluate the Determinant Expand the determinant along the first row: $1 \cdot \begin{vmatrix} 2 & -1 \\ \alpha-6 & 3 \end{vmatrix} - (-1) \cdot \begin{vmatrix} \alpha & -1 \\ 8 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} \alpha & 2 \\ 8 & \alpha-6 \end{vmatrix} = 0$ Calculate each $2 \times 2$ determinant:

• $(2)(3) - (-1)(\alpha-6) = 6 - (-\alpha+6) = 6 + \alpha - 6 = \alpha$
• $(\alpha)(3) - (-1)(8) = 3\alpha + 8$
• $(\alpha)(\alpha-6) - (2)(8) = \alpha^2 - 6\alpha - 16$

Substitute these values back into the determinant expansion: $1 \cdot (\alpha) + 1 \cdot (3\alpha + 8) + 1 \cdot (\alpha^2 - 6\alpha - 16) = 0$ $\alpha + 3\alpha + 8 + \alpha^2 - 6\alpha - 16 = 0$

Step 5: Simplify and Solve the Quadratic Equation Combine like terms to form a quadratic equation in $\alpha$: $\alpha^2 + (\alpha + 3\alpha - 6\alpha) + (8 - 16) = 0$ $\alpha^2 - 2\alpha - 8 = 0$ Factor the quadratic equation: We look for two numbers that multiply to $-8$ and add to $-2$. These numbers are $-4$ and $2$. $(\alpha - 4)(\alpha + 2) = 0$ This yields two possible values for $\alpha$:

• $\alpha - 4 = 0 \implies \alpha = 4$
• $\alpha + 2 = 0 \implies \alpha = -2$

Step 6: Calculate the Sum of All Values of $\alpha$ The problem asks for the sum of all possible values of $\alpha$. Sum $= 4 + (-2) = 2$.

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Common Mistakes & Tips

• Vector Subtraction Errors: Be meticulous when subtracting position vectors. Ensure the components are subtracted in the correct order ($\vec{OB} - \vec{OA}$).
• Determinant Calculation: Double-check the signs and products when expanding the $3 \times 3$ determinant. A common mistake is miscalculating the cofactors or their signs.
• Quadratic Equation Roots: If you form a quadratic equation, you can use Vieta's formulas to find the sum of the roots directly. For an equation $a\alpha^2 + b\alpha + c = 0$, the sum of roots is $-b/a$. In this case, for $\alpha^2 - 2\alpha - 8 = 0$, the sum of roots is $-(-2)/1 = 2$.

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Summary

The condition for four points to be coplanar is that the scalar triple product of the three vectors formed by taking one point as the origin and connecting it to the other three points must be zero. This condition translates into evaluating a $3 \times 3$ determinant, which, in this case, results in a quadratic equation in $\alpha$. Solving this quadratic equation gives the possible values of $\alpha$, and their sum is the required answer.

The final answer is $\boxed{2}$.