If `a, b, c, l, m, n in R-{0}` such that `al+bm+cn=0, bl+cm+an=0, cl+am+bn=0`. If a, b, c are distinct and `f(x)=ax^(3)+bx^(2)+cx+2`. Find f(1) .

Answer

To solve the problem, we need to find the value of \( f(1) \) given the polynomial \( f(x) = ax^3 + bx^2 + cx + 2 \) and the conditions involving \( a, b, c, l, m, n \).

### Step-by-step Solution:

1. **Understanding the Given Conditions**:
We have the following equations:
\[ al + bm + cn = 0 \]
\[ bl + cm + an = 0 \]
\[ cl + am + bn = 0 \]
These equations suggest a relationship among \( a, b, c, l, m, n \).

2. **Matrix Representation**:
We can represent these equations in matrix form:
\[ \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \begin{pmatrix} l \\ m \\ n \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
For a non-trivial solution (where \( l, m, n \) are not all zero), the determinant of the coefficient matrix must be zero.

3. **Calculating the Determinant**:
We need to calculate the determinant:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Using properties of determinants, we can perform row operations to simplify it.

4. **Row Operations**:
Let's perform the row operation \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\[ D = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Now, we can factor out \( a+b+c \):
\[ D = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]

5. **Expanding the Determinant**:
The determinant can be expanded:
\[ D = (a+b+c) \left( 1 \cdot (ca - ab) - 1 \cdot (cb - ac) + 1 \cdot (ba - bc) \right) \]
Simplifying this gives us:
\[ D = (a+b+c)(ca - ab + ba - bc + cb - ac) \]

6. **Conclusion from the Determinant**:
Since \( a, b, c \) are distinct, the only way for \( D = 0 \) is if:
\[ a + b + c = 0 \]

7. **Finding \( f(1) \)**:
Now, substituting \( x = 1 \) into the polynomial:
\[ f(1) = a(1^3) + b(1^2) + c(1) + 2 = a + b + c + 2 \]
Since we established that \( a + b + c = 0 \):
\[ f(1) = 0 + 2 = 2 \]

### Final Answer:
Thus, the value of \( f(1) \) is \( \boxed{2} \).