If `x gt m,ygt n,z gt r (x,y,z gt 0)` such that `|{:(x,n,r),(m,y,r),(m,n,z):}|=0` then the greatest value of
`(27xyz)/((x-m)(y-n)(z-r))` is

Answer

To solve the problem, we need to evaluate the determinant and find the greatest value of the expression \(\frac{27xyz}{(x-m)(y-n)(z-r)}\) given that the determinant \(|\begin{pmatrix} x & n & r \\ m & y & r \\ m & n & z \end{pmatrix}| = 0\).

### Step-by-Step Solution:

1. **Set Up the Determinant:**
We start with the determinant:
\[ D = \begin{vmatrix} x & n & r \\ m & y & r \\ m & n & z \end{vmatrix} \]
Given that \(D = 0\).

2. **Row Operations:**
We perform row operations to simplify the determinant. We can replace \(R_2\) with \(R_2 - R_3\):
\[ D = \begin{vmatrix} x & n & r \\ m & y-n & r-z \\ m & n & z \end{vmatrix} \]

3. **Expand the Determinant:**
We can expand the determinant using the first row:
\[ D = x \begin{vmatrix} y-n & r-z \\ n & z \end{vmatrix} - n \begin{vmatrix} m & r-z \\ m & z \end{vmatrix} + r \begin{vmatrix} m & y-n \\ m & n \end{vmatrix} \]

4. **Calculate the 2x2 Determinants:**
- For the first determinant:
\[ \begin{vmatrix} y-n & r-z \\ n & z \end{vmatrix} = (y-n)z - n(r-z) = yz - nz - nr + nz = yz - nr \]
- For the second determinant:
\[ \begin{vmatrix} m & r-z \\ m & z \end{vmatrix} = mz - m(r-z) = mz - mr + mz = 2mz - mr \]
- For the third determinant:
\[ \begin{vmatrix} m & y-n \\ m & n \end{vmatrix} = mn - m(y-n) = mn - my + mn = 2mn - my \]

5. **Substituting Back:**
Substitute these back into the determinant:
\[ D = x(yz - nr) - n(2mz - mr) + r(2mn - my) \]

6. **Set the Determinant to Zero:**
Since \(D = 0\), we can rearrange this equation to find a relationship between \(x\), \(y\), and \(z\).

7. **Using AM-GM Inequality:**
From the properties of determinants and the conditions \(x > m\), \(y > n\), \(z > r\), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{x}{x-m} + \frac{y}{y-n} + \frac{z}{z-r} \geq 3\sqrt[3]{\frac{xyz}{(x-m)(y-n)(z-r)}} \]

8. **Finding the Maximum Value:**
From the determinant condition, we derive that:
\[ \frac{x}{x-m} + \frac{y}{y-n} + \frac{z}{z-r} = 2 \]
Thus, we have:
\[ 2 \geq 3\sqrt[3]{\frac{xyz}{(x-m)(y-n)(z-r)}} \]

9. **Cubing Both Sides:**
Cubing both sides gives:
\[ 8 \geq \frac{27xyz}{(x-m)(y-n)(z-r)} \]

10. **Conclusion:**
Therefore, the greatest value of \(\frac{27xyz}{(x-m)(y-n)(z-r)}\) is \(8\).

### Final Answer:
The greatest value of \(\frac{27xyz}{(x-m)(y-n)(z-r)}\) is \(\boxed{8}\).