This problem requires us to combine concepts from 3D geometry, specifically the area of a triangle and the perpendicular distance from a point to a line. We are given the coordinates of one vertex, the line containing the other two vertices, the area of the triangle, and the length of its base. Our goal is to find the value of $\alpha^2$, where $\alpha$ is an integer.

1. Key Concepts and Formulas

• Area of a Triangle: The area of a triangle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ In $\triangle ABC$, if $BC$ is considered the base, then the height is the perpendicular distance from vertex $A$ to the line containing $BC$.

• Perpendicular Distance from a Point to a Line in 3D: For a point $P_A(x_0, y_0, z_0)$ and a line passing through a point $P_0(x_1, y_1, z_1)$ with a direction vector $\vec{d}=(a,b,c)$, the perpendicular distance $h$ from $P_A$ to the line is given by: $h = \frac{|\vec{P_0P_A} \times \vec{d}|}{|\vec{d}|}$ Here, $\vec{P_0P_A}$ is the vector from any point $P_0$ on the line to the point $P_A$.

2. Step-by-Step Solution

Step 1: Calculate the Height of the Triangle We are given the area of $\triangle ABC$ as 21 square units and the length of the base $BC$ as $2\sqrt{21}$ units. We use the area formula to find the height ($h$) from vertex $A$ to the line containing $BC$: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $21 = \frac{1}{2} \times (2\sqrt{21}) \times h$ $21 = \sqrt{21} \times h$ To find $h$, we divide both sides by $\sqrt{21}$: $h = \frac{21}{\sqrt{21}} = \sqrt{21}$ Thus, the perpendicular distance from vertex $A$ to the line containing $BC$ is $\sqrt{21}$ units.

Step 2: Identify a Point and Direction Vector of the Line The line containing vertices $B$ and $C$ is given in symmetric form: $\frac{x + \alpha}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ To use the perpendicular distance formula, we need a point on the line ($P_0$) and its direction vector ($\vec{d}$).

• Point on the line ($P_0$): We can find a point on the line by setting the numerators to zero. $x + \alpha = 0 \implies x = -\alpha$ $y - 1 = 0 \implies y = 1$ $z + 4 = 0 \implies z = -4$ So, a point on the line is $P_0(-\alpha, 1, -4)$.
• Direction vector of the line ($\vec{d}$): The denominators of the symmetric form give the direction ratios. So, $\vec{d} = (5, 2, 3)$.

Step 3: Calculate the Vector from a Point on the Line to Vertex A The vertex $A$ is given as $A(0, 2, \alpha)$. We need the vector $\vec{P_0A}$ (vector from $P_0$ to $A$): $\vec{P_0A} = A - P_0 = (0 - (-\alpha), 2 - 1, \alpha - (-4)) = (\alpha, 1, \alpha+4)$

Step 4: Calculate the Cross Product $\vec{P_0A} \times \vec{d}$ Now we compute the cross product of $\vec{P_0A} = (\alpha, 1, \alpha+4)$ and $\vec{d} = (5, 2, 3)$: $\vec{P_0A} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3 \end{vmatrix}$ Expanding the determinant:

• i-component: $(1)(3) - (\alpha+4)(2) = 3 - (2\alpha + 8) = -2\alpha - 5$
• j-component: $-(\alpha(3) - (\alpha+4)(5)) = -(3\alpha - 5\alpha - 20) = -(-2\alpha - 20) = 2\alpha + 20$
• k-component: $(\alpha)(2) - (1)(5) = 2\alpha - 5$ So, $\vec{P_0A} \times \vec{d} = (-2\alpha - 5, 2\alpha + 20, 2\alpha - 5)$.

Step 5: Calculate the Magnitudes of the Vectors We need $|\vec{P_0A} \times \vec{d}|$ and $|\vec{d}|$.

• Magnitude of the direction vector $|\vec{d}|$: $|\vec{d}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{25 + 4 + 9} = \sqrt{38}$
• Magnitude of the cross product $|\vec{P_0A} \times \vec{d}|$: $|\vec{P_0A} \times \vec{d}|^2 = (-2\alpha - 5)^2 + (2\alpha + 20)^2 + (2\alpha - 5)^2$ Since $(-x)^2 = x^2$, we can write $(-2\alpha - 5)^2 = (2\alpha + 5)^2$. $|\vec{P_0A} \times \vec{d}|^2 = (2\alpha + 5)^2 + (2\alpha + 20)^2 + (2\alpha - 5)^2$ Expand each term: $(4\alpha^2 + 20\alpha + 25) + (4\alpha^2 + 80\alpha + 400) + (4\alpha^2 - 20\alpha + 25)$ Combine like terms: $|\vec{P_0A} \times \vec{d}|^2 = (4\alpha^2 + 4\alpha^2 + 4\alpha^2) + (20\alpha + 80\alpha - 20\alpha) + (25 + 400 + 25)$ $|\vec{P_0A} \times \vec{d}|^2 = 12\alpha^2 + 80\alpha + 450$ Therefore, $|\vec{P_0A} \times \vec{d}| = \sqrt{12\alpha^2 + 80\alpha + 450}$.

Step 6: Apply the Perpendicular Distance Formula and Solve for $\alpha$ We know that the height $h = \sqrt{21}$. Using the formula $h = \frac{|\vec{P_0A} \times \vec{d}|}{|\vec{d}|}$: $\sqrt{21} = \frac{\sqrt{12\alpha^2 + 80\alpha + 450}}{\sqrt{38}}$ Square both sides of the equation to eliminate the square roots: $21 = \frac{12\alpha^2 + 80\alpha + 450}{38}$ Multiply both sides by 38: $21 \times 38 = 12\alpha^2 + 80\alpha + 450$ $798 = 12\alpha^2 + 80\alpha + 450$ Rearrange into a quadratic equation: $12\alpha^2 + 80\alpha + 450 - 798 = 0$ $12\alpha^2 + 80\alpha - 348 = 0$ Divide the entire equation by 4 to simplify: $3\alpha^2 + 20\alpha - 87 = 0$ Now, solve this quadratic equation for $\alpha$ using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\alpha = \frac{-20 \pm \sqrt{20^2 - 4(3)(-87)}}{2(3)}$ $\alpha = \frac{-20 \pm \sqrt{400 + 1044}}{6}$ $\alpha = \frac{-20 \pm \sqrt{1444}}{6}$ Calculate the square root of 1444: $\sqrt{1444} = 38$. $\alpha = \frac{-20 \pm 38}{6}$ This gives two possible values for $\alpha$:

1. $\alpha_1 = \frac{-20 + 38}{6} = \frac{18}{6} = 3$
2. $\alpha_2 = \frac{-20 - 38}{6} = \frac{-58}{6} = -\frac{29}{3}$

Step 7: Determine the Integer Value of $\alpha$ The problem states that $\alpha \in \mathbb{Z}$ ($\alpha$ is an integer). Comparing our two solutions:

• $\alpha = 3$ is an integer.
• $\alpha = -\frac{29}{3}$ is not an integer. Therefore, the valid value for $\alpha$ is 3.

Step 8: Calculate $\alpha^2$ Finally, we need to find $\alpha^2$: $\alpha^2 = 3^2 = 9$

3. Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with calculations, especially when expanding squared terms in the magnitude of the cross product. Sign errors are common.
• Vector Direction: Ensure consistency when forming the vector from a point on the line to vertex A (e.g., $\vec{P_0A}$ or $\vec{AP_0}$). While the magnitude will be the same, the components will be negated.
• Quadratic Formula: Double-check the values of $a, b, c$ and the calculation of the discriminant in the quadratic formula.

4. Summary

The problem was solved by first utilizing the given area and base length to determine the height of the triangle. This height represents the perpendicular distance from vertex A to the line containing BC. We then extracted a point and direction vector from the given line equation. Using the 3D formula for the perpendicular distance from a point to a line, which involves a cross product and magnitudes, we set up an equation in terms of $\alpha$. Solving this quadratic equation yielded two values for $\alpha$, from which the integer value was selected based on the problem's constraint. Finally, we calculated $\alpha^2$.

The final answer is $\boxed{9}$.