Question : If $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$, find the value of $(pa+qb+rc)$.
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
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Correct Answer: 0
Solution : Given: $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$ Let $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k$ $\frac{a}{q-r}=k$ Multiplying by $p$, we get: $pa=k(pq-pr)$ -------------------(1) Similarly, $qb=k(qr-qp)$ ------------------(2) and $rc=k(rp-rq)$ --------------------------(3) Adding equations (1), (2), and (3), we get, $⇒pa+qb+rc=k(pq-pr+qr-qp+rp-rq)$ $\therefore pa+qb+rc=0$ Hence, the correct answer is 0.
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