An easy equation.

$x(3{x}^{2} -19) = 2({x}^{2} - 11)$

$3{x}^{3} -2{x}^{2} - 19x + 22 = 0 = f(x)$

Now, we know one of the roots is in the form of $a$ , hence we will first find the value of this root. For doing this, we will use the factor-remainder theorem to check what number will be the root of this equation.

After checking by hit-and-trial,

x = 2 is a root of the equation because

$3*{2}^{3} - 2*{2}^{2} - 19*2 + 22$

24 - 8 - 38 + 22 = $0$

So, a = 2

As,

f(x) = (x - a)(x - $\frac{r + \sqrt{m}}{k}$ )(x - $\frac{r-\sqrt{m}}{k}$ )

Therefore, (x - $\frac{r + \sqrt{m}}{k}$ )(x - $\frac{r-\sqrt{m}}{k}$ ) = f(x)/(x-2)

= $3{x}^{2} + 4x -11$

Now, we will find the roots of this quadratic equation.

Hence, $\alpha$ , $\beta$ = $\frac {-4 \pm \sqrt{ {-4}^{2} - 4* 3* -11}}{2*3}$

= $\frac{-4 \pm \sqrt{148}}{6}$

= $\frac{ -2 \pm \sqrt{37}}{3}$

Therefore, a + r + m + k =2 +(-2) + 37 + 3 = $\boxed{40}$