Q250 / 270 · NDA 2017 (Phase 2) · Mathematics
If $p+q+r=a+b+c=0$, then the determinant $\left|\begin{array}{lll}p a & q b & r c \\ q c & \text { ra } & p b \\ r b & p c & q a\end{array}\right|$ equals
- Correct answerA$0$
- B$1$
- C$\mathrm{pa}+\mathrm{qb}+\mathrm{rc}$
- D$\mathrm{pa}+\mathrm{qb}+\mathrm{rc}+\mathrm{a}+\mathrm{b}+\mathrm{c}$
Solution
$\left|\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}\right|=\mathrm{pa}\left(\mathrm{rqa}^2-\mathrm{p}^2 \mathrm{bc}\right)-\mathrm{qb}\left(\mathrm{q}^2 \mathrm{ac}-\mathrm{prb}^2\right)+$
$\begin{aligned} & \operatorname{rc}\left(q \mathrm{qcc}^2-\mathrm{r}^2 a b\right) \\ & =\mathrm{pqra}^3-\mathrm{p}^3 a b c-\mathrm{q}^3 a b c+\mathrm{pqrb}^3+\mathrm{pqrc}^3-\mathrm{r}^3 a b c \\ & =\mathrm{pqr}\left(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3\right)-\mathrm{abc}\left(\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3\right) \\ & \text {Given, } \mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}=0\end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3=3 \mathrm{pqr} \text { and } \mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=3 \mathrm{abc} . \\ & \therefore \operatorname{pqr}\left(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3\right)-\mathrm{abc}\left(\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3\right) \\ & =\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr}) \\ & =0\end{aligned}$
Hence, the correct answer is option: 'A'.