If $\begin{vmatrix}p-a& 0&c-r\\0& q-b&c-r\\a&b&r\end{vmatrix}=0$ then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is

2

The correct answer is Option (3) → 2

Given:

$\left|\begin{array}{ccc} p - a & 0 & c - r \\ 0 & q - b & c - r \\ a & b & r \end{array}\right| = 0$

Expand the determinant along the second row:

$= -0 \cdot \text{minor} + (q - b) \cdot \left[(p - a)r - a(c - r)\right] - (c - r) \cdot \left[(p - a)b - 0 \right] = 0$

$\Rightarrow (q - b)\left[(p - a)r - a(c - r)\right] = (c - r)(p - a)b$

Expanding both sides:

$(q - b)(pr - ar - ac + ar) = (c - r)(pb - ab)$

$\Rightarrow (q - b)(pr - ac) = (c - r)(pb - ab)$

Now divide both sides by $(p - a)(q - b)(r - c)$:

$\Rightarrow \frac{pr - ac}{p - a} + \frac{qc - br}{q - b} + \frac{ab - pq}{r - c} = 0$

Simplifying leads to:

$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2$