Let `A(0,6,6)`, B(6,6,0) and C(6,0,6) are three points and point D is moving on the line `x+z-3=0=y`. If G is centroid of `DeltaABC`, then minimum value of GD is

Options

A. `sqrt((47)/2)`
B. `sqrt((37)/2)`
C. `sqrt((57)/2)`
D. `sqrt((23)/2)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Find the coordinates of the centroid G of triangle ABC.
The coordinates of points A, B, and C are given as:
- A(0, 6, 6)
- B(6, 6, 0)
- C(6, 0, 6)

The formula for the centroid \( G \) of a triangle with vertices \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) is:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) \]

Calculating the coordinates of G:
\[ G\left(\frac{0 + 6 + 6}{3}, \frac{6 + 6 + 0}{3}, \frac{6 + 0 + 6}{3}\right) = G\left(\frac{12}{3}, \frac{12}{3}, \frac{12}{3}\right) = G(4, 4, 4) \]

### Step 2: Determine the coordinates of point D on the line.
The line is defined by the equations:
\[ x + z - 3 = 0 \quad \text{and} \quad y = 0 \]
From the first equation, we can express \( x \) in terms of \( z \):
\[ x = 3 - z \]
Thus, the coordinates of point D can be expressed as:
\[ D(3 - z, 0, z) \]

### Step 3: Find the distance GD.
The distance \( GD \) between points \( G(4, 4, 4) \) and \( D(3 - z, 0, z) \) is given by the distance formula:
\[ GD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ GD = \sqrt{(4 - (3 - z))^2 + (4 - 0)^2 + (4 - z)^2} \]
This simplifies to:
\[ GD = \sqrt{(1 + z)^2 + 4^2 + (4 - z)^2} \]
Calculating further:
\[ GD = \sqrt{(1 + z)^2 + 16 + (4 - z)^2} \]
Expanding the squares:
\[ GD = \sqrt{(1 + 2z + z^2) + 16 + (16 - 8z + z^2)} \]
Combining like terms:
\[ GD = \sqrt{2z^2 - 6z + 33} \]

### Step 4: Minimize the distance GD.
To minimize \( GD \), we minimize \( GD^2 \):
\[ GD^2 = 2z^2 - 6z + 33 \]
This is a quadratic equation in the form \( az^2 + bz + c \), where \( a = 2 \), \( b = -6 \), and \( c = 33 \). The minimum value occurs at:
\[ z = -\frac{b}{2a} = -\frac{-6}{2 \cdot 2} = \frac{6}{4} = \frac{3}{2} \]

### Step 5: Calculate the minimum value of GD^2.
Substituting \( z = \frac{3}{2} \) back into \( GD^2 \):
\[ GD^2 = 2\left(\frac{3}{2}\right)^2 - 6\left(\frac{3}{2}\right) + 33 \]
Calculating each term:
\[ GD^2 = 2 \cdot \frac{9}{4} - 9 + 33 = \frac{18}{4} - 9 + 33 = \frac{18}{4} - \frac{36}{4} + \frac{132}{4} = \frac{114}{4} = \frac{57}{2} \]

### Step 6: Find GD.
Taking the square root gives:
\[ GD = \sqrt{\frac{57}{2}} \]

### Final Answer:
The minimum value of \( GD \) is \( \sqrt{\frac{57}{2}} \).