If the sine of the angles of `DeltaABC` satisfy the equation `c^(3)x^(3)-c^(2) (a+b+c)x^(2)+lx +m=0`
(where a,b,c are the sides of `DeltaABC),` then `DeltaABC` is

Options

A. A. always right angled for any l, m
B. B. right angled only when `l=c (ab+bc+ca) =c sum ab, m=-abc`
C. C. right angled only when `l= (c sum ab)/(4) , m=-(abc)/(8)`
D. D. never right angled

Answer

To solve the question, we need to analyze the given polynomial equation and the properties of triangle \( \Delta ABC \). The equation provided is:

\[ c^3 x^3 - c^2 (a + b + c) x^2 + l x + m = 0 \]

where \( a, b, c \) are the sides of triangle \( \Delta ABC \). We are tasked with determining the conditions under which this triangle is a right triangle.

### Step 1: Identify the Roots
The roots of the polynomial are given as \( \sin A, \sin B, \sin C \), where \( A, B, C \) are the angles of triangle \( ABC \).

### Step 2: Use Vieta's Formulas
According to Vieta's formulas, the sum of the roots can be expressed as:

\[ \sin A + \sin B + \sin C = -\frac{b}{a} \]

In our case, \( a = c^3 \) and \( b = -c^2(a + b + c) \). Therefore, we have:

\[ \sin A + \sin B + \sin C = \frac{c^2(a + b + c)}{c^3} = \frac{a + b + c}{c} \]

### Step 3: Apply the Sine Rule
From the sine rule, we know:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

where \( R \) is the circumradius of triangle \( ABC \). Thus, we can express \( \sin A, \sin B, \sin C \) as:

\[ \sin A = \frac{a}{2R}, \quad \sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R} \]

### Step 4: Substitute into the Sum of Roots
Substituting these into the sum of roots gives:

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

Equating this to our previous result:

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

### Step 5: Simplify the Equation
Assuming \( a + b + c \neq 0 \), we can cancel \( a + b + c \) from both sides:

\[ \frac{1}{2R} = \frac{1}{c} \implies 2R = c \]

### Step 6: Relate to Right Triangle Condition
From the relationship \( 2R = c \), we can deduce that:

\[ c = 2R \sin C \]

Substituting \( 2R \) gives:

\[ c = c \sin C \]

This implies:

\[ \sin C = 1 \implies C = 90^\circ \]

### Step 7: Conditions for \( l \) and \( m \)
Next, we need to find the conditions for \( l \) and \( m \) based on the product of the roots:

\[ \sin A \sin B \sin C = -\frac{m}{c^3} \]

Substituting the values of \( \sin A, \sin B, \sin C \):

\[ \frac{abc}{(2R)^3} = -\frac{m}{c^3} \]

This leads to:

\[ abc = -m \cdot \frac{8R^3}{c^3} \]

### Conclusion
From the analysis, we conclude that triangle \( \Delta ABC \) is a right triangle under the conditions:

1. \( l = c \cdot \sum ab \)
2. \( m = -abc \)

Thus, the correct option is **B**.