1. Key Concepts and Formulas

• Coordinate System for Mutually Perpendicular Edges: When three edges of a tetrahedron meeting at a vertex are mutually perpendicular, we can simplify geometric calculations by placing that vertex at the origin $(0,0,0)$ and aligning the edges with the coordinate axes.
• Area of a Triangle using Vector Cross Product: For a triangle with vertices $P, Q, R$, its area is given by $\frac{1}{2} |\vec{PQ} \times \vec{PR}|$. This formula is universally applicable for any triangle in 3D space.
• Area of a Triangle formed by Intercepts on Coordinate Axes (Pappus's Area Theorem / Projection Formula): If the vertices of a triangle $BCD$ lie on the coordinate axes (i.e., $B=(a,0,0), C=(0,b,0), D=(0,0,c)$), and the areas of the triangles formed by projecting $BCD$ onto the coordinate planes are $S_{xy} = \text{Area}(\triangle ABC)$, $S_{yz} = \text{Area}(\triangle ACD)$, and $S_{xz} = \text{Area}(\triangle ADB)$, then the area of $\triangle BCD$, denoted $S_{BCD}$, satisfies the relation: $S_{BCD}^2 = S_{xy}^2 + S_{yz}^2 + S_{xz}^2$ This provides a direct shortcut for this specific type of problem.

2. Step-by-Step Solution

Step 1: Set up a Coordinate System

• What we are doing: We are strategically placing the tetrahedron in a 3D Cartesian coordinate system to simplify calculations, leveraging the given information that edges $AB, AC, AD$ are mutually perpendicular.
• Why: The mutual perpendicularity of $AB, AC, AD$ allows us to align them with the coordinate axes. Placing vertex $A$ at the origin $(0,0,0)$ makes it straightforward to assign coordinates to the other vertices.
• Math: Let vertex $A$ be at the origin $(0,0,0)$. Let the lengths of the edges be $AB = a$, $AC = b$, and $AD = c$. Since the edges are mutually perpendicular, we can place the other vertices on the axes: $B = (a,0,0)$ $C = (0,b,0)$ $D = (0,0,c)$
• Reasoning: Assuming $a,b,c$ are positive lengths simplifies notation, as areas depend only on the magnitudes of these lengths.

Step 2: Use Given Areas to Form Equations for $a, b, c$

• What we are doing: We use the given areas of $\triangle ABC$, $\triangle ACD$, and $\triangle ADB$ to establish relationships between the lengths $a, b, c$.

• Why: Each of these triangles is a right-angled triangle at vertex $A$. Their areas can be directly calculated using the formula $\frac{1}{2} \times \text{base} \times \text{height}$.

• Math: Area of $\triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2}ab$. Given Area($\triangle ABC$) = 5. $\frac{1}{2}ab = 5 \implies ab = 10 \quad (1)$

  Area of $\triangle ACD = \frac{1}{2} \times AC \times AD = \frac{1}{2}bc$. Given Area($\triangle ACD$) = 6. $\frac{1}{2}bc = 6 \implies bc = 12 \quad (2)$

  Area of $\triangle ADB = \frac{1}{2} \times AD \times AB = \frac{1}{2}ca$. Given Area($\triangle ADB$) = 7. $\frac{1}{2}ca = 7 \implies ca = 14 \quad (3)$

• Reasoning: These equations provide the necessary products of edge lengths, which will be essential for calculating the area of $\triangle BCD$.

Step 3: Calculate Vectors for $\triangle BCD$

• What we are doing: We determine two vectors representing two sides of $\triangle BCD$, originating from a common vertex.
• Why: To apply the vector cross product formula for the area of $\triangle BCD$, we need two vectors that define its sides. Choosing vertex $C$ as the common origin for these vectors is convenient.
• Math: The vertices of $\triangle BCD$ are $B=(a,0,0)$, $C=(0,b,0)$, and $D=(0,0,c)$. Vector $\vec{CB} = B - C = (a-0, 0-b, 0-0) = (a, -b, 0)$. Vector $\vec{CD} = D - C = (0-0, 0-b, c-0) = (0, -b, c)$.
• Reasoning: These vectors will be used in the cross product calculation to find the area.

Step 4: Compute the Cross Product of the Vectors

• What we are doing: We calculate the cross product of $\vec{CB}$ and $\vec{CD}$.
• Why: The magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by these vectors. The area of the triangle formed by these vectors is half of this magnitude.
• Math: $\vec{CB} \times \vec{CD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & -b & 0 \\ 0 & -b & c \end{vmatrix}$ $= \mathbf{i}((-b)(c) - (0)(-b)) - \mathbf{j}((a)(c) - (0)(0)) + \mathbf{k}((a)(-b) - (-b)(0))$ $= \mathbf{i}(-bc) - \mathbf{j}(ac) + \mathbf{k}(-ab)$ $= (-bc, -ac, -ab)$
• Reasoning: This vector is normal to the plane containing $\triangle BCD$, and its magnitude will be used to determine the triangle's area.

Step 5: Calculate the Magnitude of the Cross Product

• What we are doing: We find the magnitude of the vector obtained from the cross product in Step 4.
• Why: The area of $\triangle BCD$ is half the magnitude of the cross product $\vec{CB} \times \vec{CD}$.
• Math: $|\vec{CB} \times \vec{CD}| = \sqrt{(-bc)^2 + (-ac)^2 + (-ab)^2}$ $= \sqrt{b^2c^2 + a^2c^2 + a^2b^2}$
• Reasoning: This expression gives us the necessary value to compute the area of $\triangle BCD$.

Step 6: Substitute Values and Calculate Area of $\triangle BCD$

• What we are doing: We substitute the squared values of $ab, bc, ca$ (derived in Step 2) into the magnitude formula to find the numerical area.

• Why: This is the final step to obtain the numerical value of the area of $\triangle BCD$.

• Math: From Step 2, we have: $ab = 10 \implies a^2b^2 = 10^2 = 100$ $bc = 12 \implies b^2c^2 = 12^2 = 144$ $ca = 14 \implies c^2a^2 = 14^2 = 196$

  Substitute these squared values into the magnitude expression: $|\vec{CB} \times \vec{CD}| = \sqrt{144 + 196 + 100}$ $= \sqrt{440}$

  Now, calculate the Area($\triangle BCD$): $\text{Area}(\triangle BCD) = \frac{1}{2} |\vec{CB} \times \vec{CD}| = \frac{1}{2} \sqrt{440}$ To simplify $\sqrt{440}$: $\sqrt{440} = \sqrt{4 \times 110} = 2\sqrt{110}$ Therefore, $\text{Area}(\triangle BCD) = \frac{1}{2} (2\sqrt{110}) = \sqrt{110}$

• Reasoning: This yields the final numerical answer for the area.

3. Common Mistakes & Tips

• Forgetting the $\frac{1}{2}$ Factor: A frequent error is to forget to divide the magnitude of the cross product by 2. The cross product's magnitude gives the area of a parallelogram, and a triangle is half of that.
• Incorrect Coordinate Assignment: Ensure that when placing vertex $A$ at the origin, the other vertices $B, C, D$ are correctly assigned to the respective coordinate axes based on which edges are mutually perpendicular.
• Using the Projection Formula (Shortcut): For problems involving a tetrahedron where three edges meeting at a vertex are mutually perpendicular, a significant shortcut is available. The square of the area of the face opposite to this vertex (e.g., $\triangle BCD$) is equal to the sum of the squares of the areas of the other three faces (which are projections onto the coordinate planes). In this case: $S_{BCD}^2 = S_{ABC}^2 + S_{ACD}^2 + S_{ADB}^2$ $S_{BCD}^2 = 5^2 + 6^2 + 7^2 = 25 + 36 + 49 = 110$ $S_{BCD} = \sqrt{110}$. This method is much faster.

4. Summary The problem asked for the area of $\triangle BCD$ in a tetrahedron where edges $AB, AC, AD$ are mutually perpendicular. We began by setting up a coordinate system with vertex $A$ at the origin and aligning $AB, AC, AD$ with the $x, y, z$ axes, respectively. This allowed us to define the coordinates of $B, C, D$ in terms of the edge lengths $a, b, c$. Using the given areas of $\triangle ABC$, $\triangle ACD$, and $\triangle ADB$, we established relationships for $ab$, $bc$, and $ca$. We then calculated the area of $\triangle BCD$ by taking half the magnitude of the cross product of two vectors representing its sides, $\vec{CB}$ and $\vec{CD}$. This calculation, substituting the squared products of edge lengths, yielded the area as $\sqrt{110}$ square units.

5. Final Answer The area of the $\triangle BCD$ is $\sqrt{110}$ square units.

The final answer is \boxed{\sqrt{110}} which corresponds to option (A).