Question:

Given the determinant equation $\begin{vmatrix} x & n & r \\ m & y & r \\ m & n & z \end{vmatrix} = 0$. <br> Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$: <br> $\begin{vmatrix} x-m & n-y & 0 \\ 0 & y-n & r-z \\ m & n & z \end{vmatrix} = 0$ <br> Expanding along the first row: <br> $(x-m)[z(y-n) - n(r-z)] - (n-y)[0 - m(r-z)] = 0$ <br> $(x-m)(y-n)z - (x-m)n(r-z) + (y-n)m(r-z) = 0$ <br> Dividing the entire equation by $(x-m)(y-n)(z-r)$: <br> $\frac{z}{z-r} - \frac{n}{y-n} + \frac{m}{x-m} = 0$ <br> $\Rightarrow \frac{z}{z-r} + \frac{m}{x-m} + \frac{n}{y-n} = 0$ <br> Since $\frac{x}{x-m} = 1 + \frac{m}{x-m}$ and $\frac{y}{y-n} = 1 + \frac{n}{y-n}$, we can rewrite the equation as: <br> $\frac{z}{z-r} + \left(\frac{x}{x-m} - 1\right) + \left(\frac{y}{y-n} - 1\right) = 0$ <br> $\Rightarrow \frac{x}{x-m} + \frac{y}{y-n} + \frac{z}{z-r} = 2$