If `(p )/(a) + (q)/(b) = (r)/(c) = 1 and (a )/(p )+ (b)/(q) + (c)/(r) = 0` where p,q,r and a,b,c are non-zero, then the value of `(p ^(2))/( a ^(2)) + (q ^(2))/(b ^(2)) + (r ^(2))/( c ^(2))` is:

Options

A. `-1`
B. `0`
C. 1
D. 2

Answer

To solve the problem, we start with the given equations:

1. \(\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\)
2. \(\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\)

We need to find the value of \(\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2}\).

### Step 1: Square the first equation
We square the first equation:
\[ \left(\frac{p}{a} + \frac{q}{b} + \frac{r}{c}\right)^2 = 1^2 = 1 \]

### Step 2: Expand the squared equation
Using the identity \((x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz)\), we can expand:
\[ \frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2} + 2\left(\frac{p}{a} \cdot \frac{q}{b} + \frac{q}{b} \cdot \frac{r}{c} + \frac{r}{c} \cdot \frac{p}{a}\right) = 1 \]

### Step 3: Substitute the second equation
From the second equation, we know:
\[ \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0 \]
This can be rearranged to:
\[ \frac{b}{q} + \frac{c}{r} = -\frac{a}{p} \]
Multiplying by \(\frac{1}{abc}\) gives:
\[ \frac{bc}{q} + \frac{ac}{r} + \frac{ab}{p} = 0 \]

### Step 4: Relate the products
Using the above relation, we can express the products:
\[ \frac{p}{a} \cdot \frac{q}{b} + \frac{q}{b} \cdot \frac{r}{c} + \frac{r}{c} \cdot \frac{p}{a} = \frac{pq}{ab} + \frac{qr}{bc} + \frac{rp}{ca} \]

### Step 5: Substitute back into the equation
Substituting this back into our expanded equation gives:
\[ \frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2} + 2 \cdot 0 = 1 \]
Thus, we have:
\[ \frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2} = 1 \]

### Conclusion
The value of \(\frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2}\) is \(1\).

### Final Answer
\[ \frac{p^2}{a^2} + \frac{q^2}{b^2} + \frac{r^2}{c^2} = 1 \]