Complete the square into the sphere equation in standard form−3x^2+36x−3y^2+14y−3z^2−30z−98=0

To complete the square and convert the given equation into the standard form of a sphere equation, we need to group the x, y, and z terms separately and complete the square for each. The standard form of a sphere equation is (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2, where (h,k,l) is the center of the sphere and r is the radius.

Given equation: -3x^2 + 36x - 3y^2 + 14y - 3z^2 - 30z - 98 = 0

First, we can factor out the coefficients of the squared terms from each group:

-3(x^2 - 12x) - 3(y^2 - 14/3*y) - 3(z^2 + 10z) = 98

Next, we complete the square for each group. This involves taking half of the coefficient of x, y, and z, squaring it and adding it to both sides of the equation:

-3[(x - 6)^2 - 36] - 3[(y - 7/3)^2 - 49/9] - 3[(z + 5)^2 - 25] = 98

Simplify the equation:

-3(x - 6)^2 + 108 - 3(y - 7/3)^2 + 49 - 3(z + 5)^2 + 75 = 98

Combine like terms:

-3(x - 6)^2 - 3(y - 7/3)^2 - 3(z + 5)^2 = 98 - 108 - 49 - 75

Simplify the right side:

-3(x - 6)^2 - 3(y - 7/3)^2 - 3(z + 5)^2 = -134

Finally, divide through by -3 to get the equation in standard form:

(x - 6)^2 + (y - 7/3)^2 + (z + 5)^2 = 134/3

So, the center of the sphere is at (6, 7/3, -5) and the radius is sqrt(134/3).