use the rational root theorem

Let $a=3y-2$ , $b=4y+4$ and $c=y+14$ . From the Heron's Formula, we have

$s=\dfrac{a+b+c}{2}=\dfrac{3y-2+4y+4+y+14}{2}=4y+8$

It follows that,

$s-a=4y-8-(3y-2)=y+10$

$s-b=4y+8-(4y+4)=4$

$s-c=4y+8-(y+14)=3y-6$

Then,

$24\sqrt{14}=\sqrt{(4y+8)(y+10)(4)(3y-6)}$

Squaring both sides we get,

$8064=(4y+8)(y+10)(4)(3y-6)$

Then simplify,

$8064=(4y^2+48y+80)(12y-24)$

$8064=48y^3-96y^2+576y^2-1152y+960y-1920$

$48y^3+480y^2-192y-9984=0$

$y^3+10y^2-4y-208$

Factor,

$(y-4)(y^2+14y+52)=0$

$y-4=0$

$y=4$

Substituting $y=4$ , we have

$a=3y-2=12-2=10$

$b=4y+4=16+4=20$

$c=y+14=18$

The desired answer is $\boxed{a=10}$ .

NOTE:

$y^2+14y+52=0$ has non-real and imaginary roots because $b^2<4ac:14^2<4(52):196<208$