If `f_r(x),g_(r)(x),h_r(x),r=1,2,3` are polynomials in x such that `f_r(a) = g_r(a) = h_r (a) , r = 1,2,3` and `F(x) = |(f_1(x),f_2(x),f_3(x)),(g_1(x),g_2(x),g_3(x)),(h_1(x),h_2(x),h_3(x))|` then F' at x = a is ………

Answer

To solve the problem, we need to find the derivative of the determinant \( F(x) \) at the point \( x = a \). The determinant is defined as:

\[ F(x) = \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \]

Given that \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), we can denote this common value as \( K_r \). Therefore, we have:

- \( f_1(a) = g_1(a) = h_1(a) = K_1 \)
- \( f_2(a) = g_2(a) = h_2(a) = K_2 \)
- \( f_3(a) = g_3(a) = h_3(a) = K_3 \)

### Step 1: Differentiate the Determinant

Using the property of determinants, the derivative of \( F(x) \) can be computed using the formula for the derivative of a determinant:

\[ F'(x) = \sum_{i=1}^{n} \det(A_i) \]

where \( A_i \) is the matrix formed by differentiating the \( i^{th} \) row of the determinant and keeping the others unchanged.

### Step 2: Construct the Derivatives

We will differentiate each row of the determinant:

1. **Differentiate the first row:**
\[ F'(x) = \begin{vmatrix} f_1'(x) & f_2'(x) & f_3'(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \]

2. **Differentiate the second row:**
\[ + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1'(x) & g_2'(x) & g_3'(x) \\ h_1(x) & h_2(x) & h_3(x) \end{vmatrix} \]

3. **Differentiate the third row:**
\[ + \begin{vmatrix} f_1(x) & f_2(x) & f_3(x) \\ g_1(x) & g_2(x) & g_3(x) \\ h_1'(x) & h_2'(x) & h_3'(x) \end{vmatrix} \]

### Step 3: Evaluate at \( x = a \)

Now we substitute \( x = a \):

\[ F'(a) = \begin{vmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1'(a) & g_2'(a) & g_3'(a) \\ h_1(a) & h_2(a) & h_3(a) \end{vmatrix} + \begin{vmatrix} f_1(a) & f_2(a) & f_3(a) \\ g_1(a) & g_2(a) & g_3(a) \\ h_1'(a) & h_2'(a) & h_3'(a) \end{vmatrix} \]

Since \( f_r(a) = g_r(a) = h_r(a) \) for \( r = 1, 2, 3 \), we have:

\[ F'(a) = \begin{vmatrix} f_1'(a) & f_2'(a) & f_3'(a) \\ K_1 & K_2 & K_3 \\ K_1 & K_2 & K_3 \end{vmatrix} + \begin{vmatrix} K_1 & K_2 & K_3 \\ g_1'(a) & g_2'(a) & g_3'(a) \\ K_1 & K_2 & K_3 \end{vmatrix} + \begin{vmatrix} K_1 & K_2 & K_3 \\ K_1 & K_2 & K_3 \\ h_1'(a) & h_2'(a) & h_3'(a) \end{vmatrix} \]

### Step 4: Simplify the Determinants

In each of these determinants, two rows are identical (the second and third rows in each case), which means that each determinant evaluates to zero. Therefore:

\[ F'(a) = 0 + 0 + 0 = 0 \]

### Final Answer

Thus, the derivative \( F'(a) \) is:

\[ \boxed{0} \]