Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $$\triangle BCD$$ is equal to:

Let the lengths of the mutually perpendicular edges $$AB$$, $$AC$$, and $$AD$$ be $$x$$, $$y$$, and $$z$$ respectively.

Step 1: Express the given face areas in terms of edge lengths

The areas of the right-angled triangles are given by:

$$\text{Area}(\triangle ABC) = \frac{1}{2}xy = 5 \implies xy = 10$$

$$\text{Area}(\triangle ACD) = \frac{1}{2}yz = 6 \implies yz = 12$$

$$\text{Area}(\triangle ADB) = \frac{1}{2}zx = 7 \implies zx = 14$$

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Step 2: Relate the tetrahedron along the 3D plane

Let vertex $$A$$ be the origin $$(0,0,0)$$. Then the coordinates of the vertices are:

$$A(0, 0, 0), \quad B(x, 0, 0), \quad C(0, y, 0), \quad D(0, 0, z)$$

We form two side vectors for triangle $$BCD$$ from vertex $$B$$:

$$\vec{BC} = -x\hat{i} + y\hat{j}$$

$$\vec{BD} = -x\hat{i} + z\hat{k}$$

The cross product $$\vec{BC} \times \vec{BD}$$ is evaluated using a determinant:

$$\vec{BC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -x & y & 0 \\ -x & 0 & z \end{vmatrix} = (yz)\hat{i} + (zx)\hat{j} + (xy)\hat{k}$$

The area of triangle $$BCD$$ is half the magnitude of this cross product vector:

$$\text{Area}(\triangle BCD) = \frac{1}{2}\sqrt{(yz)^2 + (zx)^2 + (xy)^2}$$

$$\text{Area}(\triangle BCD) = \sqrt{\left(\frac{1}{2}yz\right)^2 + \left(\frac{1}{2}zx\right)^2 + \left(\frac{1}{2}xy\right)^2}$$

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Step 3: Substitute the known area values

Substituting the given face areas into the derived relation:

$$\text{Area}(\triangle BCD) = \sqrt{6^2 + 7^2 + 5^2}$$

$$\text{Area}(\triangle BCD) = \sqrt{36 + 49 + 25} = \sqrt{110}$$