The plane `(x)/(1)+(y)/(2)+(z)/(3)=1` intersect x - axis, y - axis at A, B and C respectively. If the distance between the origin and the controid of `DeltaABC` is `k_(1)` units and the volume of the tetrahedron OABC is `k_(2)` cubic units, then the value of `(k_(1)^(2))/(k_(2))` is equal to (where O is the origin)

Options

A. 21
B. `(14)/(9)`
C. 63
D. `(14)/(3)`

Answer

To solve the problem, we need to find the coordinates of points A, B, and C where the plane intersects the x-axis, y-axis, and z-axis respectively. Then, we will calculate the centroid of triangle ABC, the distance from the origin to the centroid, and the volume of tetrahedron OABC.

### Step 1: Find the intercepts 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-intercept)**: Set \(y = 0\) and \(z = 0\):
\[ \frac{x}{1} = 1 \implies x = 1 \implies A(1, 0, 0) \]

- **Point B (y-intercept)**: Set \(x = 0\) and \(z = 0\):
\[ \frac{y}{2} = 1 \implies y = 2 \implies B(0, 2, 0) \]

- **Point C (z-intercept)**: Set \(x = 0\) and \(y = 0\):
\[ \frac{z}{3} = 1 \implies z = 3 \implies C(0, 0, 3) \]

### Step 2: Find the centroid of triangle ABC
The coordinates of 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:

\[ k_1 = \sqrt{\left(\frac{1}{3} - 0\right)^2 + \left(\frac{2}{3} - 0\right)^2 + (1 - 0)^2} \]

Calculating this:

\[ k_1 = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + 1^2} = \sqrt{\frac{1}{9} + \frac{4}{9} + 1} = \sqrt{\frac{5}{9} + 1} = \sqrt{\frac{5}{9} + \frac{9}{9}} = \sqrt{\frac{14}{9}} = \frac{\sqrt{14}}{3} \]

### Step 4: Calculate the volume of tetrahedron OABC
The volume \(k_2\) of tetrahedron OABC can be calculated using the formula:

\[ k_2 = \frac{1}{6} \left| \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \right| \]

Where:
- \(\vec{OA} = (1, 0, 0)\)
- \(\vec{OB} = (0, 2, 0)\)
- \(\vec{OC} = (0, 0, 3)\)

Calculating the cross product \(\vec{OB} \times \vec{OC}\):

\[ \vec{OB} \times \vec{OC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} = \hat{i}(2 \cdot 3 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 0) = 6\hat{i} \]

Now, calculate the dot product:

\[ \vec{OA} \cdot (\vec{OB} \times \vec{OC}) = (1, 0, 0) \cdot (6, 0, 0) = 6 \]

Thus, the volume is:

\[ k_2 = \frac{1}{6} \cdot 6 = 1 \]

### Step 5: Calculate \(\frac{k_1^2}{k_2}\)
Now we can find \(\frac{k_1^2}{k_2}\):

\[ k_1^2 = \left(\frac{\sqrt{14}}{3}\right)^2 = \frac{14}{9} \]

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:

\[ \frac{14}{9} \]