If the rectangular faces of a brick have their diagonals in the ratio $3:2\sqrt{3}:\sqrt{15}$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

Correct Option: 3 (representing $1:\sqrt{3}$)

Let the edges of the brick be $a$, $b$, and $c$ in increasing order of length, i.e., $a\le b\le c$.

The diagonals of the three rectangular faces of the brick will be:
$d_1=\sqrt{a^2+b^2}$
$d_2=\sqrt{a^2+c^2}$
$d_3=\sqrt{b^2+c^2}$

Since $a\le b\le c$, the diagonals will also be ordered as:
$\sqrt{a^2+b^2}\le \sqrt{a^2+c^2}\le \sqrt{b^2+c^2}$

Given that the ratio of the diagonals is $3:2\sqrt{3}:\sqrt{15}$, we can write the squares of the ratios to simplify:
$(a^2+b^2):(a^2+c^2):(b^2+c^2)=3^2:{\left(2\sqrt{3}\right)}^2:{\left(\sqrt{15}\right)}^2=9:12:15$

Let:
$a^2+b^2=9k$
$a^2+c^2=12k$
$b^2+c^2=15k$

Adding all three equations, we get:
$2(a^2+b^2+c^2)=36k$
$a^2+b^2+c^2=18k$

Now we can find individual values:
$c^2=(a^2+b^2+c^2)-(a^2+b^2)=18k-9k=9k$
$b^2=(a^2+b^2+c^2)-(a^2+c^2)=18k-12k=6k$
$a^2=(a^2+b^2+c^2)-(b^2+c^2)=18k-15k=3k$

Thus, the lengths of the edges are:
$a=\sqrt{3k}$
$c=\sqrt{9k}$

The shortest edge is $a$ and the longest edge is $c$ (since $3k<6k<9k$).
The ratio of the shortest edge to the longest edge is:
$\frac{a}{c}=\frac{\sqrt{3k}}{\sqrt{9k}}=\sqrt{\frac{3}{9}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$

Hence, the ratio is $1:\sqrt{3}$.