Given,

The plane $x+3y-2z+6=0$ meet the co-ordinate axes at the points $A, B, C$,

So, $A\left(-6,0,0\right), B\left(0,-2,0\right) \& C\left(0,0,3\right)$

And the orthocenter of the triangle $ABC$ is $\left(\alpha , \beta , \frac{6}{7}\right)$,

We know that, circumcentre of triangle is point of intersection of perpendicular bisectors,

So, plotting the diagram we get,

Now from diagram we can see that $O\text{ to midpoint of} BC \perp BC$

So, by perpendicular vector formula we get,

$\left(x-0\right)\times 0+\left(y+1\right)\times 2+\left(z-\frac{3}{2}\right)\times 3=0$

$\Rightarrow 4y+6z-5=0 .......\left(1\right)$

Similarly, for side $AC$ we get,

$\left(x+3\right)\times \left(-6\right)+\left(y-0\right)\times 0+\left(z-\frac{3}{2}\right)\left(-3\right)=0$

$\Rightarrow 4x+2z+9=0 .........\left(2\right)$

Similarly, for side $AB$ we get,

$\Rightarrow \left(x+3\right)\times \left(-6\right)+\left(y+1\right)\times 2+\left(z-0\right)\times 0=0$

$\Rightarrow 3x-y+8=0 .........\left(3\right)$

Now from equation $\left(1\right), \left(2\right) \& \left(3\right)$ we get,

$x=\frac{-9}{4}-\frac{z}{2}, y=\frac{5}{4}-\frac{3}{2}z, z=z$

Now using the relation between centroid, circumcentre and orthocentre, we get,

$\Rightarrow G=\left(\frac{\alpha -{\displaystyle \frac{9}{2}}-z}{3}, \frac{\beta +{\displaystyle \frac{5}{2}}-3z}{3}, \frac{{\displaystyle \frac{6}{7}}+2z}{3}\right)$

$\Rightarrow \left(-2,-\frac{2}{3},1\right)=\left(\frac{\alpha -{\displaystyle \frac{9}{2}}-z}{3}, \frac{\beta +{\displaystyle \frac{5}{2}}-3z}{3}, \frac{{\displaystyle \frac{6}{7}+2z}}{3}\right)$

Now on comparing both side we get,

$\Rightarrow z=\frac{15}{14}, \alpha =-\frac{3}{7}, \beta =-\frac{9}{7}$

$\Rightarrow 98{\left(\alpha +\beta \right)}^2=98\times \frac{144}{49}=288$