1. Set up coordinates

Since $AB, AC, AD$ are mutually perpendicular, place $A=(0,0,0),\quad B=(x,0,0),\quad C=(0,y,0),\quad D=(0,0,z)$ with $x,y,z>0$.

Then:

• $AB=x$
• $AC=y$
• $AD=z$

2. Use the given face areas

Because each of the triangles $ABC, ACD, ADB$ is right-angled at $A$,

$[ABC]=\frac12 xy=5 \implies xy=10$ $[ACD]=\frac12 yz=6 \implies yz=12$ $[ADB]=\frac12 xz=7 \implies xz=14$

3. Find $x^2,y^2,z^2$

We use $x^2=\frac{(xy)(xz)}{yz}=\frac{10\cdot 14}{12}=\frac{35}{3}$ $y^2=\frac{(xy)(yz)}{xz}=\frac{10\cdot 12}{14}=\frac{60}{7}$ $z^2=\frac{(xz)(yz)}{xy}=\frac{14\cdot 12}{10}=\frac{84}{5}$

4. Area of triangle $BCD$

Using coordinates, $\overrightarrow{BC}=(-x,y,0),\quad \overrightarrow{BD}=(-x,0,z)$

Area of $\triangle BCD$ is $\frac12\left|\overrightarrow{BC}\times \overrightarrow{BD}\right|$

Now, $\overrightarrow{BC}\times \overrightarrow{BD}=(yz,xz,xy)$ so $\left|\overrightarrow{BC}\times \overrightarrow{BD}\right|=\sqrt{(yz)^2+(xz)^2+(xy)^2}$

Hence, $[BCD]=\frac12\sqrt{(12)^2+(14)^2+(10)^2}$ $=\frac12\sqrt{144+196+100}$ $=\frac12\sqrt{440}$ $=\sqrt{110}$

5. Check options

• A: $\sqrt{110}$ ✅
• B: $12$ ❌
• C: $\sqrt{340}$ ❌
• D: $7\sqrt{3}$ ❌

Therefore, the correct answer is $\boxed{\sqrt{110}}$