Let the plane $$x + 3y - 2z + 6 = 0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$ABC$$ is $$\left(\alpha, \beta, \frac{6}{7}\right)$$, then $$98(\alpha + \beta)^2$$ is equal to _____.

The plane $$x + 3y - 2z + 6 = 0$$ meets the coordinate axes at points A, B, C. We find the orthocenter of triangle ABC.

Setting $$y = z = 0$$: $$x = -6$$, so $$A = (-6, 0, 0)$$.

Setting $$x = z = 0$$: $$3y = -6$$, so $$B = (0, -2, 0)$$.

Setting $$x = y = 0$$: $$-2z = -6$$, so $$C = (0, 0, 3)$$.

Let the orthocenter be $$H = (\alpha, \beta, \frac{6}{7})$$.

The orthocenter satisfies $$\vec{AH} \perp \vec{BC}$$ and $$\vec{BH} \perp \vec{AC}$$.

$$\vec{AH} = (\alpha + 6, \beta, \frac{6}{7})$$, $$\vec{BC} = (0, 2, 3)$$

$$0(\alpha+6) + 2\beta + 3 \cdot \frac{6}{7} = 0 \Rightarrow 2\beta + \frac{18}{7} = 0 \Rightarrow \beta = -\frac{9}{7}$$

$$\vec{BH} = (\alpha, \beta + 2, \frac{6}{7})$$, $$\vec{AC} = (6, 0, 3)$$

$$6\alpha + 0(\beta+2) + 3 \cdot \frac{6}{7} = 0 \Rightarrow 6\alpha + \frac{18}{7} = 0 \Rightarrow \alpha = -\frac{3}{7}$$

$$\alpha + \beta = -\frac{3}{7} - \frac{9}{7} = -\frac{12}{7}$$

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

The correct answer is 288.