A sphere has the equation `|r-a|^2+|r-b|^2=72, where a=hat(i)+3hat(j)-6hat(k) and b=2hat(i)+4hat(j)+2hat(k)`
Find
(i) The centre of sphere
(ii) The radius of sphere
(iii) Perpendicular distance from the centre of the sphere to the plane `rcdot(2hat(i)+2hat(j)-hat(k))+3=0`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given vectors and the equation of the sphere
We are given:
- \( a = \hat{i} + 3\hat{j} - 6\hat{k} \)
- \( b = 2\hat{i} + 4\hat{j} + 2\hat{k} \)
- The equation of the sphere is \( |r - a|^2 + |r - b|^2 = 72 \).

### Step 2: Rewrite the equation of the sphere
The equation \( |r - a|^2 + |r - b|^2 = 72 \) can be expanded as follows:
Let \( r = x\hat{i} + y\hat{j} + z\hat{k} \).

Then,
\[ |r - a|^2 = |(x - 1)\hat{i} + (y - 3)\hat{j} + (z + 6)\hat{k}|^2 = (x - 1)^2 + (y - 3)^2 + (z + 6)^2 \]
\[ |r - b|^2 = |(x - 2)\hat{i} + (y - 4)\hat{j} + (z - 2)\hat{k}|^2 = (x - 2)^2 + (y - 4)^2 + (z - 2)^2 \]

Thus, the equation becomes:
\[ (x - 1)^2 + (y - 3)^2 + (z + 6)^2 + (x - 2)^2 + (y - 4)^2 + (z - 2)^2 = 72 \]

### Step 3: Expand and simplify the equation
Expanding both squared terms:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 3)^2 = y^2 - 6y + 9 \]
\[ (z + 6)^2 = z^2 + 12z + 36 \]
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 4)^2 = y^2 - 8y + 16 \]
\[ (z - 2)^2 = z^2 - 4z + 4 \]

Combining these gives:
\[ 2x^2 - 6x + 5 + 2y^2 - 14y + 25 + 2z^2 + 8z + 40 = 72 \]

### Step 4: Combine like terms
Combining like terms:
\[ 2x^2 + 2y^2 + 2z^2 - 6x - 14y + 8z + 70 = 72 \]
This simplifies to:
\[ 2x^2 + 2y^2 + 2z^2 - 6x - 14y + 8z - 2 = 0 \]

### Step 5: Rearranging to standard form
Dividing the entire equation by 2:
\[ x^2 + y^2 + z^2 - 3x - 7y + 4z - 1 = 0 \]

### Step 6: Complete the square
Completing the square for each variable:
- For \( x \): \( x^2 - 3x \) becomes \( (x - \frac{3}{2})^2 - \frac{9}{4} \)
- For \( y \): \( y^2 - 7y \) becomes \( (y - \frac{7}{2})^2 - \frac{49}{4} \)
- For \( z \): \( z^2 + 4z \) becomes \( (z + 2)^2 - 4 \)

Substituting back:
\[ (x - \frac{3}{2})^2 + (y - \frac{7}{2})^2 + (z + 2)^2 - \frac{9}{4} - \frac{49}{4} - 4 - 1 = 0 \]
This simplifies to:
\[ (x - \frac{3}{2})^2 + (y - \frac{7}{2})^2 + (z + 2)^2 = 72 + \frac{9 + 49 + 16 + 4}{4} \]
Calculating the right side gives:
\[ = 72 + \frac{78}{4} = 72 + 19.5 = 91.5 \]

### Step 7: Identify the center and radius
From the equation:
- The center \( C \) of the sphere is \( \left( \frac{3}{2}, \frac{7}{2}, -2 \right) \)
- The radius \( r \) is \( \sqrt{91.5} \)

### Step 8: Calculate the perpendicular distance from the center to the plane
The plane equation is given by:
\[ 2x + 2y - z + 3 = 0 \]
The perpendicular distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \( A = 2, B = 2, C = -1, D = 3 \) and \( (x_0, y_0, z_0) = \left( \frac{3}{2}, \frac{7}{2}, -2 \right) \):
\[ D = \frac{|2(\frac{3}{2}) + 2(\frac{7}{2}) - (-2) + 3|}{\sqrt{2^2 + 2^2 + (-1)^2}} \]
Calculating the numerator:
\[ = \frac{|3 + 7 + 2 + 3|}{\sqrt{4 + 4 + 1}} = \frac{15}{3} = 5 \]

### Final Answers
(i) The center of the sphere is \( \left( \frac{3}{2}, \frac{7}{2}, -2 \right) \)
(ii) The radius of the sphere is \( \sqrt{91.5} \)
(iii) The perpendicular distance from the center of the sphere to the plane is \( 5 \).