Key Concepts and Formulas

• Area of a Parallelogram from Diagonals: If $\vec{d_1}$ and $\vec{d_2}$ are the diagonals of a parallelogram, its area is given by $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
• Vector Addition: To add vectors, add their corresponding components. If $\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$ and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$, then $\vec{u} + \vec{v} = (u_x+v_x)\hat{i} + (u_y+v_y)\hat{j} + (u_z+v_z)\hat{k}$.
• Cross Product: The cross product of two vectors $\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$ and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ is given by the determinant: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} = (u_y v_z - u_z v_y)\hat{i} - (u_x v_z - u_z v_x)\hat{j} + (u_x v_y - u_y v_x)\hat{k}$
• Magnitude of a Vector: The magnitude of a vector $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ is $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.

Step-by-Step Solution

Step 1: Determine the Diagonal Vectors $\vec{d_1}$ and $\vec{d_2}$ We are given the vectors $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$, $\vec{b}=-\hat{i}+\hat{k}$, and $\vec{c}=\beta \hat{j}-\hat{k}$. The diagonals of the parallelogram are given as $\vec{d_1} = \vec{a}+\vec{b}$ and $\vec{d_2} = \vec{b}+\vec{c}$. We need to find these vectors in terms of $\alpha$ and $\beta$.

• Calculate $\vec{d_1} = \vec{a}+\vec{b}$: $\vec{d_1} = (2 \hat{i}+\alpha \hat{j}+\hat{k}) + (-\hat{i}+\hat{k})$ $\vec{d_1} = (2-1)\hat{i} + (\alpha+0)\hat{j} + (1+1)\hat{k}$ $\vec{d_1} = \hat{i} + \alpha \hat{j} + 2\hat{k}$

• Calculate $\vec{d_2} = \vec{b}+\vec{c}$: $\vec{d_2} = (-\hat{i}+\hat{k}) + (\beta \hat{j}-\hat{k})$ $\vec{d_2} = (-1+0)\hat{i} + (0+\beta)\hat{j} + (1-1)\hat{k}$ $\vec{d_2} = -\hat{i} + \beta \hat{j}$

Step 2: Compute the Cross Product of the Diagonals, $\vec{d_1} \times \vec{d_2}$ The area of the parallelogram depends on the cross product of its diagonals. We will compute $\vec{d_1} \times \vec{d_2}$ using the determinant formula.

$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0 \end{vmatrix}$ Expanding the determinant: $\vec{d_1} \times \vec{d_2} = \hat{i}(\alpha \cdot 0 - 2 \cdot \beta) - \hat{j}(1 \cdot 0 - 2 \cdot (-1)) + \hat{k}(1 \cdot \beta - \alpha \cdot (-1))$ $\vec{d_1} \times \vec{d_2} = \hat{i}(0 - 2\beta) - \hat{j}(0 + 2) + \hat{k}(\beta + \alpha)$ $\vec{d_1} \times \vec{d_2} = -2\beta \hat{i} - 2\hat{j} + (\alpha+\beta)\hat{k}$

Step 3: Find the Magnitude of the Cross Product, $|\vec{d_1} \times \vec{d_2}|$ The area formula requires the magnitude of the cross product vector.

$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-2\beta)^2 + (-2)^2 + (\alpha+\beta)^2}$ $|\vec{d_1} \times \vec{d_2}| = \sqrt{4\beta^2 + 4 + (\alpha+\beta)^2}$

Step 4: Formulate and Solve the Area Equation We are given that the area of the parallelogram is $\frac{\sqrt{21}}{2}$. Using the area formula, we can set up an equation to find a relationship between $\alpha$ and $\beta$.

$\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$ $\frac{\sqrt{21}}{2} = \frac{1}{2} \sqrt{4\beta^2 + 4 + (\alpha+\beta)^2}$ Multiplying both sides by 2: $\sqrt{21} = \sqrt{4\beta^2 + 4 + (\alpha+\beta)^2}$ Squaring both sides: $21 = 4\beta^2 + 4 + (\alpha+\beta)^2$ $21 - 4 = 4\beta^2 + \alpha^2 + 2\alpha\beta + \beta^2$ $17 = \alpha^2 + 5\beta^2 + 2\alpha\beta$ We are also given that $\alpha$ and $\beta$ are integers and $\alpha \beta = -6$. Substituting this into the equation: $17 = \alpha^2 + 5\beta^2 + 2(-6)$ $17 = \alpha^2 + 5\beta^2 - 12$ $17 + 12 = \alpha^2 + 5\beta^2$ $29 = \alpha^2 + 5\beta^2$

Step 5: Determine the Integer Pairs $(\alpha, \beta)$ We need to find integer pairs $(\alpha, \beta)$ that satisfy both $\alpha \beta = -6$ and $29 = \alpha^2 + 5\beta^2$. The possible integer pairs for $\alpha \beta = -6$ are: $(1, -6), (-1, 6), (2, -3), (-2, 3), (3, -2), (-3, 2), (6, -1), (-6, 1)$.

Let's test each pair in the equation $29 = \alpha^2 + 5\beta^2$:

• For $(1, -6)$: $1^2 + 5(-6)^2 = 1 + 5(36) = 1 + 180 = 181 \neq 29$.
• For $(-1, 6)$: $(-1)^2 + 5(6)^2 = 1 + 5(36) = 1 + 180 = 181 \neq 29$.
• For $(2, -3)$: $2^2 + 5(-3)^2 = 4 + 5(9) = 4 + 45 = 49 \neq 29$.
• For $(-2, 3)$: $(-2)^2 + 5(3)^2 = 4 + 5(9) = 4 + 45 = 49 \neq 29$.
• For $(3, -2)$: $3^2 + 5(-2)^2 = 9 + 5(4) = 9 + 20 = 29$. This pair satisfies the equation.
• For $(-3, 2)$: $(-3)^2 + 5(2)^2 = 9 + 5(4) = 9 + 20 = 29$. This pair also satisfies the equation.
• For $(6, -1)$: $6^2 + 5(-1)^2 = 36 + 5(1) = 36 + 5 = 41 \neq 29$.
• For $(-6, 1)$: $(-6)^2 + 5(1)^2 = 36 + 5(1) = 36 + 5 = 41 \neq 29$.

The two ordered pairs $(\alpha, \beta)$ that satisfy all conditions are $(3, -2)$ and $(-3, 2)$. Let $(\alpha_1, \beta_1) = (3, -2)$ and $(\alpha_2, \beta_2) = (-3, 2)$.

Step 6: Calculate the Final Expression $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ We need to compute the value of $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ using the identified pairs.

$\alpha_1^2+\beta_1^2-\alpha_2 \beta_2 = (3)^2 + (-2)^2 - ((-3) \cdot 2)$ $= 9 + 4 - (-6)$ $= 13 + 6$ $= 19$

If we swap the assignments: $(\alpha_1, \beta_1) = (-3, 2)$ and $(\alpha_2, \beta_2) = (3, -2)$: $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2 = (-3)^2 + (2)^2 - ((3) \cdot (-2))$ $= 9 + 4 - (-6)$ $= 13 + 6$ $= 19$ The result remains the same.

Common Mistakes & Tips

• Cross Product Calculation: Be extremely careful with signs when computing the determinant for the cross product. A single sign error can lead to an incorrect result.
• Algebraic Simplification: When squaring $(\alpha+\beta)^2$, expand it fully as $\alpha^2 + 2\alpha\beta + \beta^2$. Errors in this expansion are common.
• Integer Pair Testing: Systematically test all possible integer factor pairs of $\alpha \beta = -6$ to ensure no valid pairs are missed and no invalid ones are incorrectly included.

Summary The problem required us to find the integer values of $\alpha$ and $\beta$ that satisfy conditions related to the area of a parallelogram formed by vectors derived from $\vec{a}$, $\vec{b}$, and $\vec{c}$. We first calculated the diagonal vectors, then their cross product and its magnitude. Using the given area, we derived an equation relating $\alpha$ and $\beta$. By combining this equation with the condition $\alpha \beta = -6$ and the fact that $\alpha, \beta$ are integers, we identified the two valid ordered pairs $(\alpha, \beta)$. Finally, we substituted these pairs into the expression $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ to find the required value.

The final answer is $\boxed{19}$ which corresponds to option (C).