Finding the dimensions of a cuboid given the volume, surface area, and diagonal 
Let the volume, surface area and length of the diagonal of a cuboid be as follows: $144$, $192$, $13$. Find the dimensions.

My trial: $$lbh = 144$$
$$2(lb+bh+lh) = 192 \implies lb+bh+lh = 96$$
$$l^2 + b^2 + h^2 = 13^2 = 169.$$
As $(l+b+h)^2 = l^2 + b^2 + h^2 + 2(lb+bh+lh) = 169 + 192 = 361$. So, $l+b+h = 19$.
Then?
NB: Sorry to all for posting my 1st question in a wrong manner. Thanks for helping me.

$l,b$ and $h$ are the roots of cubic euqation equation :
$x^3+bx^2+cx+d=0$
$lbh= \dfrac{-d}{1}$
$lb+bh+hl=\dfrac{c}{1}$
$l+b+h= \dfrac{-b}{1}$
Plug in the values and use Vieta's formula.
Aliter:
$lbh=144$
$l^2+b^2+h^2=169$
$169 \equiv 1 \mod 4$ 
$l^2 \equiv 0 \mod 4, b^2 \equiv 0 \mod 4 $ and $h^2 \equiv 1 \mod 4$
$ \implies$ $$l=2k$$
$$b=2q$$
$$h=2m-1$$ 
$lbh=144$
One of the three numbers are odd. Odd divisors of $144$ are $3$ and $1$, if $h=1$, $l+b \neq 19$ (Why?). Therefore, $h=3$.
$l^2+b^2=160$ ,$lb=48$ and $l+b=16$ 
$(l-b)^2=l^2+b^2-2lh= 64 \implies l-b=8$
We get $l=12$, $b=4$ and $h=3$.
