The position vectors of the four angular points of a tetrahedron OABC are `(0, 0, 0); (0, 0,2) , (0, 4,0)` and `(6, 0, 0)` respectively. A point P inside the tetrahedron is at the same distance `r` from the four plane faces of the tetrahedron. Find the value of `3r`

Explanation

To solve the problem, we need to find the distance \( r \) from a point \( P \) inside the tetrahedron \( OABC \) to the four plane faces of the tetrahedron, and then compute \( 3r \).

### Step-by-Step Solution:

1. **Identify the vertices of the tetrahedron**:
The vertices are given as:
- \( O(0, 0, 0) \)
- \( A(0, 0, 2) \)
- \( B(0, 4, 0) \)
- \( C(6, 0, 0) \)

2. **Determine the equation of the plane \( ABC \)**:
To find the equation of the plane passing through points \( A, B, C \), we can use the determinant method. The general equation of a plane can be expressed as:
\[ ax + by + cz = d \]
We can find the normal vector to the plane by using the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

- \( \overrightarrow{AB} = B - A = (0, 4, 0) - (0, 0, 2) = (0, 4, -2) \)
- \( \overrightarrow{AC} = C - A = (6, 0, 0) - (0, 0, 2) = (6, 0, -2) \)

Now, calculate the cross product:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & -2 \\ 6 & 0 & -2 \end{vmatrix} \]

This gives:
\[ \hat{i}(4 \cdot -2 - 0 \cdot -2) - \hat{j}(0 \cdot -2 - 6 \cdot -2) + \hat{k}(0 \cdot 0 - 6 \cdot 4) \]
\[ = \hat{i}(-8) - \hat{j}(12) - \hat{k}(24) \]
So, the normal vector is \( (-8, -12, -24) \) or simplified \( (2, 3, 6) \).

The equation of the plane can be written as:
\[ 2x + 3y + 6z = d \]
To find \( d \), we can substitute one of the points, say \( A(0, 0, 2) \):
\[ 2(0) + 3(0) + 6(2) = d \implies d = 12 \]
Thus, the equation of the plane \( ABC \) is:
\[ 2x + 3y + 6z = 12 \]

3. **Find the distance from point \( P(r, r, r) \) to the plane \( ABC \)**:
The distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz = D \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 - d|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \( P(r, r, r) \) into the distance formula:
\[ D = \frac{|2r + 3r + 6r - 12|}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{|11r - 12|}{\sqrt{49}} = \frac{|11r - 12|}{7} \]

4. **Set the distances from point \( P \) to the other planes equal to \( r \)**:
Since \( P \) is at the same distance \( r \) from all four planes, we can set:
\[ \frac{|11r - 12|}{7} = r \]
This leads to two cases to solve:
- Case 1: \( 11r - 12 = 7r \)
- Case 2: \( 11r - 12 = -7r \)

**Case 1**:
\[ 11r - 7r = 12 \implies 4r = 12 \implies r = 3 \]

**Case 2**:
\[ 11r + 7r = 12 \implies 18r = 12 \implies r = \frac{2}{3} \]

5. **Evaluate \( 3r \)**:
We need to find \( 3r \). Since \( r = \frac{2}{3} \) is the only valid solution (as \( r = 3 \) does not satisfy the condition of being inside the tetrahedron):
\[ 3r = 3 \times \frac{2}{3} = 2 \]

### Final Answer:
Thus, the value of \( 3r \) is \( \boxed{2} \).