Consider a triangle ABC, where c,y,z are the length of perpendicular drawn from the vertices of the triangle to the opposite sides a,b, c respectively. Let the letters `R,r S,Delta` denote the circumradius, inradius semi-perimeter and area of the triangle respectively.
If `(bx)/(c) +(cy)/(a)+(az)/(b) =(a^(2) +b^(2)+c ^(2))/(k),` then the value of k is

Options

A. R
B. S
C. 2R
D. `3/2R`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Understand the Problem
We have a triangle ABC with sides \( a, b, c \) opposite to vertices \( A, B, C \) respectively. The lengths of the perpendiculars from the vertices to the opposite sides are denoted as \( x, y, z \). We need to evaluate the expression:

\[ \frac{bx}{c} + \frac{cy}{a} + \frac{az}{b} = \frac{a^2 + b^2 + c^2}{k} \]

and find the value of \( k \).

### Step 2: Use the Formulas for Perpendiculars
The lengths of the perpendiculars from the vertices to the opposite sides can be expressed in terms of the area \( \Delta \) of the triangle:

\[ x = \frac{2\Delta}{a}, \quad y = \frac{2\Delta}{b}, \quad z = \frac{2\Delta}{c} \]

### Step 3: Substitute the Values of \( x, y, z \)
Substituting these values into the expression:

\[ \frac{b \cdot \frac{2\Delta}{a}}{c} + \frac{c \cdot \frac{2\Delta}{b}}{a} + \frac{a \cdot \frac{2\Delta}{c}}{b} \]

This simplifies to:

\[ \frac{2b\Delta}{ac} + \frac{2c\Delta}{ab} + \frac{2a\Delta}{bc} \]

### Step 4: Factor Out Common Terms
Factoring out \( 2\Delta \):

\[ 2\Delta \left( \frac{b}{ac} + \frac{c}{ab} + \frac{a}{bc} \right) \]

### Step 5: Find a Common Denominator
Finding a common denominator for the terms inside the parentheses:

\[ \frac{b^2 + c^2 + a^2}{abc} \]

Thus, we have:

\[ 2\Delta \cdot \frac{b^2 + c^2 + a^2}{abc} \]

### Step 6: Substitute the Area \( \Delta \)
We know that the area \( \Delta \) can be expressed as:

\[ \Delta = \frac{abc}{4R} \]

Substituting this into our expression gives:

\[ 2 \cdot \frac{abc}{4R} \cdot \frac{b^2 + c^2 + a^2}{abc} \]

### Step 7: Simplify the Expression
The \( abc \) terms cancel out:

\[ \frac{2(b^2 + c^2 + a^2)}{4R} = \frac{b^2 + c^2 + a^2}{2R} \]

### Step 8: Compare with Given Expression
Now we compare this with the given expression:

\[ \frac{b^2 + c^2 + a^2}{2R} = \frac{a^2 + b^2 + c^2}{k} \]

From this, we can see that:

\[ k = 2R \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{2R} \]