If a≠p,b≠q,c≠r and pbcp+aq+b2cabr=0 then pp−a+qq−b+rr−c=

pbcaqcabr=0R1→R1−R2,R2→R2−R5(p−a)[r(q−b)−b(c−r)]−(b−q)[−a(c−r)]=0r(p−a)(q−b)+b(p−a)(r−c)+a(q−b)(r−c)=0⇒rr−c+bq−b+ap−a=0⇒rr−c=−bq−b−ap−a Now, consider pp−a+qq−b+rr−c=pp−a+qq−b−bq−b−ap−a=1+1=2

If a≠p,b≠q,c≠r and pbcp+aq+b2cabr=0 then pp−a+qq−b+rr−c=

pbcaqcabr=0R1→R1−R2,R2→R2−R5(p−a)[r(q−b)−b(c−r)]−(b−q)[−a(c−r)]=0r(p−a)(q−b)+b(p−a)(r−c)+a(q−b)(r−c)=0⇒rr−c+bq−b+ap−a=0⇒rr−c=−bq−b−ap−a Now, consider pp−a+qq−b+rr−c=pp−a+qq−b−bq−b−ap−a=1+1=2

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If $a \neq p, b \neq q, c \neq r and |\begin{array}{ccc} p & b & c \\ p + a & q + b & 2 c \\ a & b & r \end{array}| = 0$ then $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} =$

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Detailed Solution

$\begin{array}{l} |\begin{array}{lll} p & b & c \\ a & q & c \\ a & b & r \end{array}| = 0 \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{5} \end{array}$
$\begin{array}{l} (p - a) [r (q - b) - b (c - r)] - (b - q) [- a (c - r)] = 0 \\ r (p - a) (q - b) + b (p - a) (r - c) + a (q - b) (r - c) = 0 \\ \Rightarrow \frac{r}{r - c} + \frac{b}{q - b} + \frac{a}{p - a} = 0 \\ \Rightarrow \frac{r}{r - c} = - \frac{b}{q - b} - \frac{a}{p - a} \end{array}$
$Now, consider \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$
$\begin{array}{l} = \frac{p}{p - a} + \frac{q}{q - b} - \frac{b}{q - b} - \frac{a}{p - a} \\ = 1 + 1 = 2 \end{array}$