Let `A` be a `3xx3` matrix given by `A=(a_(ij))_(3xx3)`. If for every column vector `X` satisfies `X'AX=0` and `a_(12)=2008`, `a_(13)=2010` and `a_(23)=-2012`. Then the value of `a_(21)+a_(31)+a_(32)=`

Options

A. `-6`
B. `2006`
C. `-2006`
D. `0`

Explanation

To solve the problem, we need to analyze the given conditions and find the values of \( a_{21} + a_{31} + a_{32} \).

### Step 1: Understand the given condition
We are given that for every column vector \( X \), the condition \( X'AX = 0 \) holds. This implies that the quadratic form defined by the matrix \( A \) is zero for all vectors \( X \).

### Step 2: Set up the matrix
Let \( A \) be represented as follows:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \]
We know that:
- \( a_{12} = 2008 \)
- \( a_{13} = 2010 \)
- \( a_{23} = -2012 \)

### Step 3: Expand the expression \( X'AX \)
Let \( X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \). Then,
\[ X'AX = \begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \]
This results in:
\[ X'AX = a_{11}x_1^2 + a_{22}x_2^2 + a_{33}x_3^2 + 2a_{12}x_1x_2 + 2a_{13}x_1x_3 + 2a_{23}x_2x_3 \]
Since \( X'AX = 0 \) for all \( X \), the coefficients of each term must equal zero.

### Step 4: Set up equations from the coefficients
From the expansion, we have:
1. \( a_{11} = 0 \)
2. \( a_{22} = 0 \)
3. \( a_{33} = 0 \)
4. \( a_{12} + a_{21} = 0 \)
5. \( a_{13} + a_{31} = 0 \)
6. \( a_{23} + a_{32} = 0 \)

### Step 5: Solve for \( a_{21}, a_{31}, a_{32} \)
Using the known values:
- From \( a_{12} + a_{21} = 0 \):
\[ 2008 + a_{21} = 0 \implies a_{21} = -2008 \]
- From \( a_{13} + a_{31} = 0 \):
\[ 2010 + a_{31} = 0 \implies a_{31} = -2010 \]
- From \( a_{23} + a_{32} = 0 \):
\[ -2012 + a_{32} = 0 \implies a_{32} = 2012 \]

### Step 6: Calculate \( a_{21} + a_{31} + a_{32} \)
Now, we compute:
\[ a_{21} + a_{31} + a_{32} = -2008 - 2010 + 2012 \]
Calculating this gives:
\[ = -2008 - 2010 + 2012 = -2006 \]

### Final Answer
Thus, the value of \( a_{21} + a_{31} + a_{32} \) is:
\[ \boxed{-2006} \]