Let the plane x + 3y - 2z + 6 = 0 meet the co-ordinate axes at the points A, B, C. If the orthocente

Q: Why can't we just find the equations of the altitude lines and intersect them to find the orthocenter?

In 3D geometry, finding the intersection of two lines is often more complex than it seems because lines might be skew. While you could technically find the equations of two altitudes and solve for their intersection, it is algebraically tedious and prone to error. The Euler Line property ($H = 3G - 2O$) is a much more powerful and efficient tool for this specific problem.

Q: Wait, why does the circumcenter $O$ have to lie on the plane of the triangle?

This is a fundamental geometric truth! The circumcenter of a triangle is the center of the circle that passes through all three vertices. Since all three vertices $A, B, C$ lie on the plane $x + 3y - 2z + 6 = 0$, the entire circumcircle must also lie on that same plane. Therefore, the center of that circle—the circumcenter—must also lie on the plane.

Q: Is the Euler Line property valid for triangles in 3D space, or is it just for 2D?

The Euler Line property is a universal property of triangles. Whether the triangle is sitting on a flat sheet of paper or floating in 3D space, the centroid $G$, circumcenter $O$, and orthocenter $H$ are always collinear, and $G$ always divides $HO$ in a $2:1$ ratio. It is a beautiful, invariant geometric truth!

Question

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 $(\alpha, \beta, \frac{6}{7})$, then $98(\alpha + \beta)^2$ is equal to __________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room where the walls are the coordinate planes. A mysterious plane, defined by the equation $x + 3y - 2z + 6 = 0$, slices through this space, creating a triangle $ABC$ where it intersects the axes.

Our goal is to find the orthocenter $H$ of this triangle using the properties of the Euler Line.

### Phase 1: The Intercepts

First, we must ground ourselves by finding where the plane touches the axes. To find the $x$-intercept $A$, we set $y=0$ and $z=0$ in the equation $x + 3y - 2z + 6 = 0$, which gives $x = -6$. Thus, $A = (-6, 0, 0)$.

Repeating this for $B$ (the $y$-axis) and $C$ (the $z$-axis), we find:
$$B = (0, -2, 0)$$
$$C = (0, 0, 3)$$

### Phase 2: The Centroid

The centroid $G$ is the balance point of the triangle. It is calculated as the average of the vertices:

$$G = \left( \frac{-6+0+0}{3}, \frac{0-2+0}{3}, \frac{0+0+3}{3} ight) = \left( -2, -\frac{2}{3}, 1 ight)$$

This point serves as the essential anchor for our Euler Line calculation.

### Phase 3: The Elusive Circumcenter

Next, we determine the circumcenter $O(x, y, z)$. This point is equidistant from $A, B,$ and $C$, satisfying $OA^2 = OB^2$ and $OB^2 = OC^2$.

Expanding these distance equations, the squared terms cancel out, leaving us with the following linear system:
$$3x - y + 8 = 0$$
$$4y + 6z - 5 = 0$$

Since $O$ must also lie on the plane $x + 3y - 2z + 6 = 0$, we solve this system of three linear equations to find:
$$O = \left( -\frac{39}{14}, -\frac{5}{14}, \frac{15}{14} ight)$$

### Phase 4: The Euler Line Masterclass

We utilize the elegant Euler Line property, which states that the orthocenter $H$, centroid $G$, and circumcenter $O$ are collinear such that $H = 3G - 2O$.

We calculate the coordinates of $H = (\alpha, \beta, \gamma)$ as follows:
$$\alpha = 3(-2) - 2\left( -\frac{39}{14} ight) = -6 + \frac{39}{7} = -\frac{3}{7}$$
$$\beta = 3\left( -\frac{2}{3} ight) - 2\left( -\frac{5}{14} ight) = -2 + \frac{5}{7} = -\frac{9}{7}$$

### Phase 5: The Final Victory

We are tasked with calculating the value of $98(\alpha + \beta)^2$. First, we find the sum:
$$\alpha + \beta = -\frac{3}{7} - \frac{9}{7} = -\frac{12}{7}$$

Squaring this result gives:
$$\left( -\frac{12}{7} ight)^2 = \frac{144}{49}$$

Finally, we compute the result:
$$98 	imes \frac{144}{49} = 2 	imes 144 = 288$$

The final answer is **288**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the plane $x + 3 y - 2 z + 6 = 0$ meet the co-ordinate axes at the points $A, B, C$ . If the orthocenter of the triangle $A B C$ is $(α, β, \frac{6}{7})$ , then $98 (α + β)^{2}$ is equal to __________.

Enter Numerical Value:

Visualized Solution

Plane Equation and Intercepts

Finding Intercepts $A, B, C$

Calculating Centroid $G$

Circumcenter $O$ Equidistance Property

Second Equidistance Relation

Circumcenter $O$ Planar Constraint

Solving for Circumcenter $O$

The Euler Line Relation

Solving for $α$

Solving for $β$

Final Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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