If `(Px)/((b-c))=(Qy)/((c-a))=(Rz)/((a-b))`, then find `Pax+Qby+Rcz`.

Options

A. `P+Q+R`
B. `x+y+z`
C. `a+b+c`
D. `0`

Explanation

To solve the equation given in the question, we start with the expression:

\[ \frac{Px}{b-c} = \frac{Qy}{c-a} = \frac{Rz}{a-b} \]

Let’s denote the common ratio as \( k \). Therefore, we can express \( Px \), \( Qy \), and \( Rz \) in terms of \( k \):

1. From the first equation:
\[ Px = k(b - c) \]

2. From the second equation:
\[ Qy = k(c - a) \]

3. From the third equation:
\[ Rz = k(a - b) \]

Now, we need to find the value of \( Pax + Qby + Rcz \):

\[ Pax + Qby + Rcz = P \cdot ax + Q \cdot by + R \cdot cz \]

Substituting the values of \( Px \), \( Qy \), and \( Rz \):

\[ Pax + Qby + Rcz = a(Px) + b(Qy) + c(Rz) \]

Substituting the expressions we derived:

\[ = a(k(b - c)) + b(k(c - a)) + c(k(a - b)) \]

Now, factor out \( k \):

\[ = k[a(b - c) + b(c - a) + c(a - b)] \]

Next, we simplify the expression inside the brackets:

1. Distributing \( a \):
\[ ab - ac \]

2. Distributing \( b \):
\[ bc - ab \]

3. Distributing \( c \):
\[ ca - cb \]

Putting it all together, we have:

\[ ab - ac + bc - ab + ca - cb \]

Now, notice that \( ab \) and \( -ab \) cancel each other out:

\[ -ab + ab + bc - cb - ac + ca = -ac + ca + bc - cb \]

This simplifies to:

\[ 0 \]

Thus, we have:

\[ Pax + Qby + Rcz = k \cdot 0 = 0 \]

Finally, the answer is:

\[ \boxed{0} \]