Key Concepts and Formulas

• Coplanarity of Four Points: Four points are coplanar if they lie in the same plane. This condition can be checked by forming three vectors from one of the points to the other three. These three vectors must then be coplanar.
• Scalar Triple Product: For three vectors $\vec{u} = u_1\widehat i + u_2\widehat j + u_3\widehat k$, $\vec{v} = v_1\widehat i + v_2\widehat j + v_3\widehat k$, and $\vec{w} = w_1\widehat i + w_2\widehat j + w_3\widehat k$, their scalar triple product is given by $[\vec{u} \quad \vec{v} \quad \vec{w}] = \left| \begin{matrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{matrix} \right|$.
• Condition for Coplanarity using Scalar Triple Product: Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if and only if their scalar triple product is zero, i.e., $[\vec{u} \quad \vec{v} \quad \vec{w}] = 0$.

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Step-by-Step Solution

Step 1: Represent the given points by their position vectors. Let the four given points be $P_1, P_2, P_3,$ and $P_4$, with their position vectors given as: $\vec{p_1} = 3\widehat i - 4\widehat j + 2\widehat k$ $\vec{p_2} = \widehat i + 2\widehat j - \widehat k$ $\vec{p_3} = -2\widehat i - \widehat j + 3\widehat k$ $\vec{p_4} = 5\widehat i - 2\alpha \widehat j + 4\widehat k$

Step 2: Form three vectors from a common origin. For the points to be coplanar, the three vectors formed by choosing one point as the common origin and extending to the other three must be coplanar. Let's choose $P_1$ as the common origin. We form the vectors $\vec{P_1P_2}$, $\vec{P_1P_3}$, and $\vec{P_1P_4}$. The vector $\vec{P_1P_2}$ is given by $\vec{p_2} - \vec{p_1}$: $\vec{P_1P_2} = (\widehat i + 2\widehat j - \widehat k) - (3\widehat i - 4\widehat j + 2\widehat k)$ $\vec{P_1P_2} = (1-3)\widehat i + (2-(-4))\widehat j + (-1-2)\widehat k$ $\vec{P_1P_2} = -2\widehat i + 6\widehat j - 3\widehat k$

The vector $\vec{P_1P_3}$ is given by $\vec{p_3} - \vec{p_1}$: $\vec{P_1P_3} = (-2\widehat i - \widehat j + 3\widehat k) - (3\widehat i - 4\widehat j + 2\widehat k)$ $\vec{P_1P_3} = (-2-3)\widehat i + (-1-(-4))\widehat j + (3-2)\widehat k$ $\vec{P_1P_3} = -5\widehat i + 3\widehat j + \widehat k$

The vector $\vec{P_1P_4}$ is given by $\vec{p_4} - \vec{p_1}$: $\vec{P_1P_4} = (5\widehat i - 2\alpha \widehat j + 4\widehat k) - (3\widehat i - 4\widehat j + 2\widehat k)$ $\vec{P_1P_4} = (5-3)\widehat i + (-2\alpha - (-4))\widehat j + (4-2)\widehat k$ $\vec{P_1P_4} = 2\widehat i + (-2\alpha + 4)\widehat j + 2\widehat k$

Step 3: Apply the condition for coplanarity. 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. This means their scalar triple product must be zero. $[\vec{P_1P_2} \quad \vec{P_1P_3} \quad \vec{P_1P_4}] = \left| \begin{matrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & -2\alpha+4 & 2 \end{matrix} \right| = 0$

Step 4: Evaluate the determinant. We expand the determinant along the first row: $-2 \left| \begin{matrix} 3 & 1 \\ -2\alpha+4 & 2 \end{matrix} \right| - 6 \left| \begin{matrix} -5 & 1 \\ 2 & 2 \end{matrix} \right| + (-3) \left| \begin{matrix} -5 & 3 \\ 2 & -2\alpha+4 \end{matrix} \right| = 0$ Calculate the $2 \times 2$ determinants:

• First term: $-2[(3)(2) - (1)(-2\alpha+4)] = -2[6 - (-2\alpha+4)] = -2[6 + 2\alpha - 4] = -2[2\alpha + 2] = -4\alpha - 4$.
• Second term: $-6[(-5)(2) - (1)(2)] = -6[-10 - 2] = -6[-12] = 72$.
• Third term: $-3[(-5)(-2\alpha+4) - (3)(2)] = -3[10\alpha - 20 - 6] = -3[10\alpha - 26] = -30\alpha + 78$.

Substitute these values back into the equation: $(-4\alpha - 4) + 72 + (-30\alpha + 78) = 0$

Step 5: Solve for $\alpha$. Combine the terms involving $\alpha$ and the constant terms: $(-4\alpha - 30\alpha) + (-4 + 72 + 78) = 0$ $-34\alpha + 146 = 0$ Add $34\alpha$ to both sides: $146 = 34\alpha$ Divide by 34: $\alpha = \frac{146}{34}$ Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2: $\alpha = \frac{146 \div 2}{34 \div 2} = \frac{73}{17}$

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Common Mistakes & Tips

• Sign Errors in Vector Subtraction: Be extremely careful when subtracting coordinates, especially when dealing with negative values. For example, $y_2 - y_1$ can become a sum if $y_1$ is negative.
• Determinant Calculation: Ensure correct expansion of the determinant and accurate arithmetic. The signs in the cofactor expansion ($+ - +$ for the first row) are critical.
• Choosing a Common Origin: Consistently use the same point as the origin for all three vectors. Mixing origins (e.g., $\vec{AB}, \vec{BC}, \vec{CD}$) will lead to an incorrect result.

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Summary

The problem requires the application of the scalar triple product to determine the coplanarity of four points. We form three vectors from a common origin and set their scalar triple product to zero. This leads to a determinant equation which, upon evaluation and simplification, yields the value of $\alpha$. The calculated value of $\alpha$ is $\frac{73}{17}$.

The final answer is $\boxed{\text{A}}$.