Moment of Inertia of Triangular loop made of rods only about an axis passing through one of the vertex and in the plane of loop parallel to the base This question is from electrodynamics although I am stuck in the mechanics part here.

$AB$,$BC$,$CA$ are wires/rods, we needed to calculate Moment of Inertia(MOI) about an axis $xx'$ parallel to BC and in the plane of loop, each wire/rod is of mass $m$ and length $l$.
I tried calculating MOI about the centroid i.e center of mass(COM), and then using parallel axes theorem, the distance between side and centroid is d = $\frac{l}{2 \sqrt3}$ so calculating MOI perpendicular to the plane about COM would be 3($\frac{ml^2}{12}$ + $md^2$). Now using perpendicular axes theoram $I_x$ + $I_y$ = $I_z$, here I know $I_y$ would be equals $\frac {ml^2}{12}$ but how do I calculate $I_x$?
Should I use this approach or is there any other approach for this?
The answer given is $\frac{5ml^2}{4}$.
Any help would be appreciated
 A: 
Look at the Fig.
The moment of Inertia towards the y' axes (green arrow) must be transformed to (x,y,z) this can obtain with this equation
$$\mathbf I_{xyz}=\mathbf R\,\mathbf I_{x'y'z'}\,\mathbf R^T$$
where  $~\mathbf R~$ isthe transformation matrix between $~x'y'z'~$ and $~xyz~$ system and $~\mathbf I~$ is the inertia tensor
$$\mathbf R= \left[ \begin {array}{ccc} \cos \left( \alpha \right) &\sin \left( 
\alpha \right) &0\\-\sin \left( \alpha \right) &
\cos \left( \alpha \right) &0\\ 0&0&1\end {array}
 \right] 
\quad,\alpha=\frac{\pi}{3}\\
\mathbf I_{x'y'z'}=\left[ \begin {array}{ccc} 0&0&0\\ 0&{\it I_{y'}}&0
\\ 0&0&0\end {array} \right]\quad\Rightarrow\\
\mathbf I_{xyz}=\left[ \begin {array}{ccc}  \left( \sin \left( \alpha \right) 
 \right) ^{2}{\it I_{y'}}&\sin \left( \alpha \right) {\it I_{y'}}\,\cos
 \left( \alpha \right) &0\\  \sin \left( \alpha
 \right) {\it I_{y'}}\,\cos \left( \alpha \right) & \left( \cos \left( 
\alpha \right)  \right) ^{2}{\it I_{y'}}&0\\  0&0&0
\end {array} \right] 
$$
thus  $~I_x=\sin^2(\alpha)\,I_{y'}=\frac{3}{4}\,\underbrace{\frac{m\,L^2}{12}}_{I_{y'}}~$
for the 3 rods you obtain
$$I_\color{red}X=2\,\left(I_{x_1}+m\,H{_1}^2\right)+m\,H_3^2\\
H_1=\frac L2\,\sin(\pi/3)\\
H_3=L\,\sin(\pi/3)$$
$\Rightarrow~I_\color{red}X=\frac 54 m\,L^2$
A: UPDATE(Final solution):
We can use the basic $\int r^2 dm$ to calculate MOI of a rod through its end point about an axis at an angle $\theta$ , which comes out to be $\frac{1}{3} ML^2 \sin^2(\theta)$, The angle $x'AC$ and $x'AB$ equals $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ respectively so we can calculate MOI of $AB$ and $AC$, now we can consider BC to be a normal point mass concentrated at the COM i.e mid point of BC and calculate its MOI using $\lambda^2 M$ where $\lambda$ is the distance of A from COM of BC and now adding MOI due to all three rods we get the final result.
You can also refer to the Tensors and rotation matrix approach given by @Eli
