To find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0), and C(6, 0, 0), we will follow these steps:
### Step 1: Find the coordinates of the midpoints of the sides of the triangle.
1. **Midpoint D of BC**:
\[
D = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}, \frac{z_B + z_C}{2}\right) = \left(\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}\right) = (3, 2, 0)
\]
2. **Midpoint E of AC**:
\[
E = \left(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}, \frac{z_A + z_C}{2}\right) = \left(\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}\right) = (3, 0, 3)
\]
3. **Midpoint F of AB**:
\[
F = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2}\right) = \left(\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}\right) = (0, 2, 3)
\]
### Step 2: Calculate the lengths of the medians.
1. **Length of Median AD**:
\[
AD = \sqrt{(x_A - x_D)^2 + (y_A - y_D)^2 + (z_A - z_D)^2}
\]
Substituting the coordinates:
\[
AD = \sqrt{(0 - 3)^2 + (0 - 2)^2 + (6 - 0)^2} = \sqrt{(-3)^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7
\]
2. **Length of Median BE**:
\[
BE = \sqrt{(x_B - x_E)^2 + (y_B - y_E)^2 + (z_B - z_E)^2}
\]
Substituting the coordinates:
\[
BE = \sqrt{(0 - 3)^2 + (4 - 0)^2 + (0 - 3)^2} = \sqrt{(-3)^2 + 4^2 + (-3)^2} = \sqrt{9 + 16 + 9} = \sqrt{34}
\]
3. **Length of Median CF**:
\[
CF = \sqrt{(x_C - x_F)^2 + (y_C - y_F)^2 + (z_C - z_F)^2}
\]
Substituting the coordinates:
\[
CF = \sqrt{(6 - 0)^2 + (0 - 2)^2 + (0 - 3)^2} = \sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7
\]
### Final Results:
- Length of median AD = 7
- Length of median BE = \(\sqrt{34}\)
- Length of median CF = 7
