1. Key Concepts and Formulas

   • Intercept Form of a Plane: A plane with equation $Ax+By+Cz+D=0$ can be rewritten in intercept form as $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a = -D/A$, $b = -D/B$, and $c = -D/C$ are the x, y, and z-intercepts respectively. These intercepts define the points where the plane meets the coordinate axes: $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$.
   • Orthocenter of a Triangle in 3D: The orthocenter $H$ of a triangle is the point where its three altitudes intersect. An altitude from a vertex is a line segment perpendicular to the opposite side (or its extension). In 3D vector geometry, if $H$ is the orthocenter of $\triangle ABC$, then the vector conditions $\vec{AH} \cdot \vec{BC} = 0$ and $\vec{BH} \cdot \vec{AC} = 0$ must hold.
   • Special Property for Axis-Intercept Triangles: For a triangle $ABC$ whose vertices lie on the coordinate axes at $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$, the orthocenter $H(x_H, y_H, z_H)$ has a specific property: it is the foot of the perpendicular from the origin $(0,0,0)$ to the plane containing $\triangle ABC$. Consequently, its coordinates satisfy $ax_H = by_H = cz_H$. Also, the orthocenter lies on the plane of the triangle, so $\frac{x_H}{a} + \frac{y_H}{b} + \frac{z_H}{c} = 1$. Combining these, the coordinates of the orthocenter are given by: $x_H = \frac{1}{a \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)}$ $y_H = \frac{1}{b \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)}$ $z_H = \frac{1}{c \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)}$

2. Step-by-Step Solution

   Step 1: Determine the coordinates of the vertices A, B, and C. The given plane equation is $x+3y-2z+6=0$. To find the intercepts, we rewrite it in intercept form: $x+3y-2z = -6$ Divide by $-6$: $\frac{x}{-6} + \frac{3y}{-6} - \frac{2z}{-6} = 1$ $\frac{x}{-6} + \frac{y}{-2} + \frac{z}{3} = 1$ Comparing this with $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, we find the intercepts: $a = -6$, $b = -2$, $c = 3$. Therefore, the coordinates of the points where the plane meets the coordinate axes are: $A(-6, 0, 0)$ $B(0, -2, 0)$ $C(0, 0, 3)$

   Step 2: Use the orthocenter property for axis-intercept triangles. Let the orthocenter of $\triangle ABC$ be $H(\alpha, \beta, \gamma)$. We are given that $\gamma = \frac{6}{7}$. According to the special property, the orthocenter $H(x_H, y_H, z_H)$ satisfies $ax_H = by_H = cz_H$. Let this common value be $k$. So, $(-6)\alpha = (-2)\beta = (3)\gamma = k$. Substitute the known value of $\gamma$: $k = (3)\left(\frac{6}{7}\right) = \frac{18}{7}$.

   Step 3: Calculate $\alpha$ and $\beta$ using the orthocenter property. Now we can find $\alpha$ and $\beta$: $(-6)\alpha = k \implies -6\alpha = \frac{18}{7} \implies \alpha = \frac{18}{7 \times (-6)} = -\frac{3}{7}$. $(-2)\beta = k \implies -2\beta = \frac{18}{7} \implies \beta = \frac{18}{7 \times (-2)} = -\frac{9}{7}$.

   Self-check using the general formula: First, calculate the denominator term: $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{(-6)^2} + \frac{1}{(-2)^2} + \frac{1}{3^2} = \frac{1}{36} + \frac{1}{4} + \frac{1}{9}$ To sum these fractions, find a common denominator, which is 36: $= \frac{1}{36} + \frac{9}{36} + \frac{4}{36} = \frac{1+9+4}{36} = \frac{14}{36} = \frac{7}{18}$. Now, calculate $\alpha, \beta, \gamma$: $\alpha = \frac{1}{a \left(\frac{7}{18}\right)} = \frac{1}{-6 \times \frac{7}{18}} = \frac{1}{-\frac{7}{3}} = -\frac{3}{7}$. $\beta = \frac{1}{b \left(\frac{7}{18}\right)} = \frac{1}{-2 \times \frac{7}{18}} = \frac{1}{-\frac{7}{9}} = -\frac{9}{7}$. $\gamma = \frac{1}{c \left(\frac{7}{18}\right)} = \frac{1}{3 \times \frac{7}{18}} = \frac{1}{\frac{7}{6}} = \frac{6}{7}$. These calculations confirm that $\alpha = -\frac{3}{7}$, $\beta = -\frac{9}{7}$, and the given $\gamma = \frac{6}{7}$ is consistent.

   Step 4: Verify that the orthocenter lies on the plane (optional but good practice). Substitute $H(-\frac{3}{7}, -\frac{9}{7}, \frac{6}{7})$ into the plane equation $x+3y-2z+6=0$: $-\frac{3}{7} + 3\left(-\frac{9}{7}\right) - 2\left(\frac{6}{7}\right) + 6$ $= -\frac{3}{7} - \frac{27}{7} - \frac{12}{7} + 6$ $= \frac{-3-27-12}{7} + 6 = \frac{-42}{7} + 6 = -6 + 6 = 0$ The orthocenter lies on the plane, confirming our values.

   Step 5: Calculate the final expression $98(\alpha+\beta)^2$. First, find $\alpha+\beta$: $\alpha+\beta = -\frac{3}{7} + \left(-\frac{9}{7}\right) = -\frac{3+9}{7} = -\frac{12}{7}$ Next, calculate $(\alpha+\beta)^2$: $(\alpha+\beta)^2 = \left(-\frac{12}{7}\right)^2 = \frac{144}{49}$ Finally, calculate $98(\alpha+\beta)^2$: $98(\alpha+\beta)^2 = 98 \times \frac{144}{49}$ $= (2 \times 49) \times \frac{144}{49} = 2 \times 144 = 288$

3. Common Mistakes & Tips

   • Incorrect Intercepts: A common error is to miscalculate the intercepts, especially with negative constants or coefficients. Always convert the plane equation to the standard intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ carefully.
   • Misapplication of Orthocenter Property: Remember that for a triangle whose vertices lie on the coordinate axes, its orthocenter is the foot of the perpendicular from the origin to the plane containing the triangle. This simplifies finding the orthocenter significantly compared to general 3D triangles.
   • Arithmetic Errors: Pay close attention to signs and fraction arithmetic, especially when squaring negative values or multiplying fractions. Double-check calculations involving fractions and common denominators.

4. Summary

   We first converted the plane equation into intercept form to find the coordinates of the vertices A, B, and C on the coordinate axes. Then, we utilized the specific property that for such a triangle, its orthocenter is the foot of the perpendicular from the origin to the plane. This allowed us to directly calculate the coordinates $\alpha = -\frac{3}{7}$ and $\beta = -\frac{9}{7}$, consistent with the given $z$-coordinate. Finally, we substituted these values into the expression $98(\alpha+\beta)^2$ to obtain the result 288.

The final answer is $\boxed{6}$.