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

Given the determinant:

$\begin{vmatrix} p-a & 0 & c-r \\ 0 & q-b & c-r \\ a & b & r \end{vmatrix} = 0$

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Applying the row operation $R_3 \to R_3 + R_1 + R_2$:

$\begin{vmatrix} p-a & 0 & c-r \\ 0 & q-b & c-r \\ a + (p-a) & b + (q-b) & r + (c-r) + (c-r) \end{vmatrix} = 0$

$\begin{vmatrix} p-a & 0 & c-r \\ 0 & q-b & c-r \\ p & q & 2c-r \end{vmatrix} = 0$

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Expanding the determinant along the first row:

$(p-a) \begin{vmatrix} q-b & c-r \\ q & 2c-r \end{vmatrix} - 0 + (c-r) \begin{vmatrix} 0 & q-b \\ p & q \end{vmatrix} = 0$

$(p-a) [(q-b)(2c-r) - q(c-r)] + (c-r) [0 - p(q-b)] = 0$

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Divide the entire equation by $(p-a)(q-b)(c-r)$:

$\frac{(p-a)[(q-b)(2c-r) - q(c-r)]}{(p-a)(q-b)(c-r)} + \frac{(c-r)[-p(q-b)]}{(p-a)(q-b)(c-r)} = 0$

$\frac{2c-r}{c-r} - \frac{q}{q-b} - \frac{p}{p-a} = 0$

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Rewrite the first term:

$\frac{c + (c-r)}{c-r} - \frac{q}{q-b} - \frac{p}{p-a} = 0$

$\frac{c}{c-r} + 1 - \frac{q}{q-b} - \frac{p}{p-a} = 0$

Rearranging the terms to match the required expression:

$\frac{p}{p-a} + \frac{q}{q-b} - \frac{c}{c-r} = 1$

Since $-\frac{c}{c-r} = \frac{c}{r-c}$, we substitute:

$\frac{p}{p-a} + \frac{q}{q-b} + \frac{c}{r-c} = 1$

We need the value of $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$. Notice that:

$\frac{r}{r-c} = \frac{(r-c)+c}{r-c} = 1 + \frac{c}{r-c}$

Substitute this back into our expression:

$\frac{p}{p-a} + \frac{q}{q-b} + \left( \frac{r}{r-c} - 1 \right) = 1$

$\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 1 + 1$

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

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The answer is Option 3: 2.