the plane x+2y+z = 6 cuts x,y,z axes at A,B,C . (alpha,beta,gamma) is the circumcentre of triangle ABC then alpha+gamma=

To find the value of alpha + gamma, where (alpha, beta, gamma) is the circumcenter of triangle ABC formed by the intersections of the plane x + 2y + z = 6 with the x, y, and z axes, we first need to determine the coordinates of points A, B, and C.

Finding the Intercepts

The plane equation is given as x + 2y + z = 6. We can find the intercepts by setting two variables to zero at a time.

Intercept on the x-axis (A)

To find point A, set y = 0 and z = 0:

• x + 2(0) + 0 = 6
• x = 6

Thus, point A is (6, 0, 0).

Intercept on the y-axis (B)

Next, for point B, set x = 0 and z = 0:

• 0 + 2y + 0 = 6
• 2y = 6
• y = 3

So, point B is (0, 3, 0).

Intercept on the z-axis (C)

Finally, for point C, set x = 0 and y = 0:

• 0 + 2(0) + z = 6
• z = 6

Therefore, point C is (0, 0, 6).

Coordinates of Triangle ABC

Now we have the vertices of triangle ABC:

• A (6, 0, 0)
• B (0, 3, 0)
• C (0, 0, 6)

Finding the Circumcenter

The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect. For triangle ABC, we can find the midpoints of two sides and then determine the equations of the perpendicular bisectors.

Midpoints

Let's calculate the midpoints:

• Midpoint of AB: M1 = ((6 + 0)/2, (0 + 3)/2, (0 + 0)/2) = (3, 1.5, 0)
• Midpoint of AC: M2 = ((6 + 0)/2, (0 + 0)/2, (0 + 6)/2) = (3, 0, 3)

Perpendicular Bisectors

Next, we find the slopes of the lines AB and AC:

• AB slope: (3 - 0) / (0 - 6) = -0.5
• AC slope: (0 - 0) / (0 - 6) = 0 (vertical line)

The perpendicular bisector of AB will have a slope of 2 (the negative reciprocal of -0.5). The equation of the line through M1 (3, 1.5) with slope 2 is:

• y - 1.5 = 2(x - 3)
• y = 2x - 6 + 1.5
• y = 2x - 4.5

The perpendicular bisector of AC is a vertical line through M2, which is x = 3.

Finding the Circumcenter Coordinates

To find the circumcenter, we solve the equations:

• x = 3
• y = 2(3) - 4.5 = 6 - 4.5 = 1.5

Thus, the circumcenter (alpha, beta, gamma) is (3, 1.5, z). To find gamma, we can substitute back into the plane equation:

• 3 + 2(1.5) + z = 6
• 3 + 3 + z = 6
• z = 0

So, the circumcenter is (3, 1.5, 0).

Final Calculation

Now, we can find alpha + gamma:

• alpha = 3
• gamma = 0

Therefore, alpha + gamma = 3 + 0 = 3.

In summary, the value of alpha + gamma is 3.