In an equilateral triangle $\text{PQR}$, side $\text{PQ}$ is divided into four equal parts, side $\text{QR}$ is divided into six equal parts and side $\text{PR}$ is divided into eight equals parts. The length of each subdivided part in $\text{cm}$ is an integer. The minimum area of the triangle $\text{PQR}$ possible, in $\text{cm}^{2}$, is

Let the side length of an equilateral triangle be $’x’\;\text{cm}.$

As mentioned in the question, the length of each subdivided part in $\text{cm}$ is an integer. So, the side length must be the LCM of $(4,6,8) \implies x = 24\;\text{cm}.$

Now, the area of an equilateral triangle of side $x\;\text{cm} = \dfrac{\sqrt{3}}{4}\; x^{2} = \dfrac{\sqrt{3}}{4} \;24^{2} = 144\;\sqrt{3}\;\text{cm}^{2}.$