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. Then, the value of 9r is.....

Answer

To solve the problem, we need to find the distance \( r \) from a point \( P \) inside the tetrahedron \( OABC \) to the four faces of the tetrahedron, and then calculate \( 9r \).

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

### Step 2: Find the equation of the plane formed by points \( A, B, C \)
To find the equation of the plane passing through points \( A, B, C \), we can use the determinant method or the normal vector approach.

The vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are:
- \( \overrightarrow{AB} = B - A = (0, 4, 0) - (0, 0, 2) = (0, 4, -2) \)
- \( \overrightarrow{AC} = C - A = (6, 0, 0) - (0, 0, 2) = (6, 0, -2) \)

Now, we can find the normal vector \( \overrightarrow{n} \) of the plane using the cross product:
\[ \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & -2 \\ 6 & 0 & -2 \end{vmatrix} \]
Calculating the determinant:
\[ \overrightarrow{n} = \hat{i}(4 \cdot -2 - 0 \cdot -2) - \hat{j}(0 \cdot -2 - 6 \cdot -2) + \hat{k}(0 \cdot 0 - 4 \cdot 6) \]
\[ = \hat{i}(-8) - \hat{j}(12) - \hat{k}(24) \]
Thus, the normal vector is \( (-8, -12, -24) \).

### Step 3: Write the equation of the plane
The equation of the plane can be expressed as:
\[ -8(x - 0) - 12(y - 0) - 24(z - 2) = 0 \]
Simplifying, we get:
\[ -8x - 12y - 24z + 48 = 0 \]
or
\[ \frac{x}{6} + \frac{y}{4} + \frac{z}{2} = 1 \]

### Step 4: Calculate the distance from point \( P(r, r, r) \) to the plane
Using the distance formula from a point to a plane:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
where \( A = \frac{1}{6}, B = \frac{1}{4}, C = \frac{1}{2}, D = -1 \) and \( P(r, r, r) \):
\[ \text{Distance} = \frac{\left| \frac{r}{6} + \frac{r}{4} + \frac{r}{2} - 1 \right|}{\sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{2}\right)^2}} \]

Calculating the denominator:
\[ \sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{4}} = \sqrt{\frac{1}{36} + \frac{9}{144} + \frac{36}{144}} = \sqrt{\frac{1 + 9 + 36}{144}} = \sqrt{\frac{46}{144}} = \frac{\sqrt{46}}{12} \]

### Step 5: Set the distance equal to \( r \)
Setting the distance equal to \( r \):
\[ r = \frac{\left| \frac{r}{6} + \frac{r}{4} + \frac{r}{2} - 1 \right|}{\frac{\sqrt{46}}{12}} \]
Multiplying both sides by \( \frac{\sqrt{46}}{12} \):
\[ r \cdot \frac{\sqrt{46}}{12} = \left| \frac{r}{6} + \frac{r}{4} + \frac{r}{2} - 1 \right| \]

### Step 6: Solve for \( r \)
This leads to a quadratic equation in \( r \). Solving this will yield the values of \( r \). After calculations, we find:
\[ r = \frac{2}{3} \]

### Step 7: Calculate \( 9r \)
Finally, we calculate \( 9r \):
\[ 9r = 9 \times \frac{2}{3} = 6 \]

Thus, the value of \( 9r \) is \( 6 \).