If \(\begin{vmatrix}p-a&0&c-r\\0&q-b&c-r\\a&b&r\end{vmatrix}=0\) then the value of \(\displaystyle\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Expanding the determinant along the first row:
\( (p-a)\big[(q-b)r - b(c-r)\big] - 0 + (c-r)\big[0 \cdot b - a(q-b)\big] = 0 \)
\( (p-a)(q-b)r - (p-a)(c-r)b - (c-r)(q-b)a = 0 \)
The fractions in the question need \(p \ne a\), \(q \ne b\) and \(r \ne c\), so dividing by \((p-a)(q-b)(c-r)\):
\( \displaystyle \frac{r}{c-r} - \frac{b}{q-b} - \frac{a}{p-a} = 0 \)
\( \displaystyle \frac{a}{p-a} + \frac{b}{q-b} = \frac{r}{c-r} = -\frac{r}{r-c} \)
Since \(\frac{p}{p-a} = 1 + \frac{a}{p-a}\) and \(\frac{q}{q-b} = 1 + \frac{b}{q-b}\),
\( \displaystyle \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} = 2 + \frac{a}{p-a} + \frac{b}{q-b} + \frac{r}{r-c} \)
\( \displaystyle = 2 - \frac{r}{r-c} + \frac{r}{r-c} = 2 \)