To solve the problem, we need to find the coordinates of the points where the plane intersects the axes, calculate the centroid of triangle ABC formed by these points, and then find the volume of the tetrahedron OABC. Finally, we will compute the value of \( \frac{k_1^2}{k_2} \).
### Step 1: Find the coordinates of points A, B, and C
The equation of the plane is given by:
\[
\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1
\]
**Point A (x-axis intersection):**
To find point A, set \( y = 0 \) and \( z = 0 \):
\[
\frac{x}{1} + 0 + 0 = 1 \implies x = 1
\]
Thus, the coordinates of point A are \( A(1, 0, 0) \).
**Point B (y-axis intersection):**
To find point B, set \( x = 0 \) and \( z = 0 \):
\[
0 + \frac{y}{2} + 0 = 1 \implies y = 2
\]
Thus, the coordinates of point B are \( B(0, 2, 0) \).
**Point C (z-axis intersection):**
To find point C, set \( x = 0 \) and \( y = 0 \):
\[
0 + 0 + \frac{z}{3} = 1 \implies z = 3
\]
Thus, the coordinates of point C are \( C(0, 0, 3) \).
### Step 2: Find the centroid of triangle ABC
The centroid \( G \) of triangle ABC can be calculated using the formula:
\[
G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)
\]
Substituting the coordinates of points A, B, and C:
\[
G\left( \frac{1 + 0 + 0}{3}, \frac{0 + 2 + 0}{3}, \frac{0 + 0 + 3}{3} \right) = G\left( \frac{1}{3}, \frac{2}{3}, 1 \right)
\]
### Step 3: Calculate the distance from the origin to the centroid
The distance \( k_1 \) from the origin \( O(0, 0, 0) \) to the centroid \( G\left( \frac{1}{3}, \frac{2}{3}, 1 \right) \) is given by:
\[
k_1 = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + 1^2}
\]
Calculating this:
\[
k_1 = \sqrt{\frac{1}{9} + \frac{4}{9} + 1} = \sqrt{\frac{1 + 4 + 9}{9}} = \sqrt{\frac{14}{9}} = \frac{\sqrt{14}}{3}
\]
### Step 4: Calculate the volume of tetrahedron OABC
The volume \( k_2 \) of tetrahedron OABC can be calculated using the formula:
\[
V = \frac{1}{6} \left| \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \right|
\]
Where:
- \( \vec{OA} = (1, 0, 0) \)
- \( \vec{OB} = (0, 2, 0) \)
- \( \vec{OC} = (0, 0, 3) \)
First, calculate the cross product \( \vec{OB} \times \vec{OC} \):
\[
\vec{OB} \times \vec{OC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & 0 \\
0 & 0 & 3
\end{vmatrix} = (2 \cdot 3) \hat{i} - (0) \hat{j} + (0) \hat{k} = 6 \hat{i}
\]
Now, calculate the dot product:
\[
\vec{OA} \cdot (\vec{OB} \times \vec{OC}) = (1, 0, 0) \cdot (6, 0, 0) = 6
\]
Thus, the volume \( k_2 \) is:
\[
k_2 = \frac{1}{6} \cdot 6 = 1
\]
### Step 5: Calculate \( \frac{k_1^2}{k_2} \)
Now, we can compute:
\[
k_1^2 = \left(\frac{\sqrt{14}}{3}\right)^2 = \frac{14}{9}
\]
Thus,
\[
\frac{k_1^2}{k_2} = \frac{\frac{14}{9}}{1} = \frac{14}{9}
\]
### Final Answer
The value of \( \frac{k_1^2}{k_2} \) is:
\[
\boxed{\frac{14}{9}}
\]
