Suppose A is a $3\times 3$ matrix. If for every column vector $X,X^{T}AX=0$ and $a_{23}=-1008,$ then the sum of the digits of $a_{32}$ is:

Use the properties of transpose of a matrix
Given, $X^{T}AX=0$ and $a_{23}=-1008$  
Let A be any $3\times 3$ square matrix, such that $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ 
Then $X^{T}AX={\left(X^{T}AX\right)}^T=X^T{A}^{T}X=-X^{T}AX\Rightarrow X^{T}AX=0$ 
Therefore, $X^T{A}^{T}X=-X^{T}AX$ 
 $A^T=-A$
$a_{23}=-a_{32}$  Or $a_{32}=-a_{23}$
 $a_{32}=-(-1008)=1008$
 Therefore , the sum of digits $=1+0+0+8=9$ 
Hence, the sum of digits of $a_{32}$ is 9