Answer
$15,18,21$ units
Work Step by Step
By Heron's formula, the area $K$ can be computed $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}.$
Hence here
$s=\dfrac{5x+6x+7x}{2}=9x$
Thus,
$K=\sqrt{9x(9x-5x)(9x-6x)(9x-7x)}
\\K=\sqrt{9x(4x)(3x)(2x)}
\\K=\sqrt{216x^4}
\\K=6\sqrt6{\space x^2}$
Therefore
$6\sqrt6\space x^2=54\sqrt6$, solve the equation above to obtain:
\begin{align*}
\dfrac{6\sqrt6\space x^2}{6\sqrt6}&=\dfrac{54\sqrt6}{6\sqrt6}\\
x^2&=9\\
x&=\pm3\end{align*}
Since $x$ represents a length, it cannot be negative so $x=3$.
Therefore, the length of the sides are:
$5x=5(3)=15\\
6x=6(3)=18\\
7x=7(3)=21$
