Wait...what????

By the Rational Root Theorem , if the roots of the polynomial $x^3 - 9x^2 + 26x - 24$ are rational, they'll be in the set $(\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24 )$ . Checking it manually, it's possible to see that $2$ , $3$ and $4$ are roots.

So, the smallest root is $2$ and the condition of the semi-perimeter gives:

$\large \displaystyle \frac{466}{2} - x = 2$

$\color{#20A900} \boxed { \large \displaystyle x = 231 }$

The equation of the perimeter is, then:

$\large \displaystyle x + y + z = 466$

$\large \displaystyle y + z = 235$

The condition of the ratio between the area and the square root of the perimeter, using the triangle's area general formula gives:

$\large \displaystyle \frac{ \frac14 \sqrt{ \left ( x+y+z \right ) \cdot \left ( x+y-z \right ) \cdot \left ( x-y+z \right ) \cdot \left ( -x+y+z \right ) } } {\sqrt{ \left ( x+y+z \right )}} = \sqrt{12834}$

The perimeter inside the numerator will cancel out with the denominator:

$\large \displaystyle \sqrt{ \left ( x+y-z \right ) \cdot \left ( x-y+z \right ) \cdot \left ( -x+y+z \right ) } = 4 \sqrt{12834}$

The terms inside the square root can be rewritten as:

$\large \displaystyle \sqrt{ \left ( x+y+z - 2z \right ) \cdot \left ( x+y+z - 2y \right ) \cdot \left ( x+y+z - 2x \right ) } = 4 \sqrt{12834}$

$\large \displaystyle \sqrt{ \left ( 466 - 2z \right ) \cdot \left ( 466 - 2y \right ) \cdot ( 4 ) } = 4 \sqrt{12834}$

$\large \displaystyle \left ( 466 - 2z \right ) \cdot \left ( 466 - 2y \right ) \cdot 4 = 16 \cdot 12834$

$\large \displaystyle 217156 - 466 \cdot 2 \cdot (y+z) + 4yz = 51336$

Since $y + z = 235$ :

$\large \displaystyle yz = 13300$

Solving:

$\large \displaystyle yz = 13300$

$\large \displaystyle y + z = 235$

One gets:

$\color{#20A900} \boxed{ \large \displaystyle y = 140}, \boxed{ \large \displaystyle z = 95}$

Or:

$\color{#20A900} \boxed{ \large \displaystyle y = 95}, \boxed{ \large \displaystyle z = 140}$

Solving for the first condition ( $y= 140$ , $z = 95$ ):

$\large \displaystyle -abc= 231$

$\large \displaystyle 10(a+b+c) = 140$

$\large \displaystyle 20(ab+ac+bc) = 95$

One gets six possible solutions:

$\color{#20A900} \boxed{ \large \displaystyle a = -3.5, b = 5.5, c = 12}$

$\color{#20A900} \boxed{ \large \displaystyle a = -3.5, b = 12, c = 5.5}$

$\color{#20A900} \boxed{ \large \displaystyle a = 5.5, b = -3.5, c = 12}$

$\color{#20A900} \boxed{ \large \displaystyle a = 5.5, b = 12, c = -3.5}$

$\color{#20A900} \boxed{ \large \displaystyle a = 12, b = -3.5, c = 5.5}$

$\color{#20A900} \boxed{ \large \displaystyle a = 12, b = 5.5, c = -3.5}$

Solving for the second condition ( $y= 95$ , $z = 140$ ), $a$ , $b$ and $c$ will have non-real solutions.

So, the only possible solutions for $a$ , $b$ and $c$ are the ones with:

$\color{#3D99F6} \boxed{ \large \displaystyle x=231, y=140, z = 95}$

Also, $(a,b,c)$ will be permutations of the set $(12, 5.5, -3.5 )$

And the digit sum of $a$ , $b$ and $c$ in any of the six cases, i.e., in any of the permutations is:

$\color{#3D99F6} \boxed{ 3+5+5+5+1+2 = 21}$