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Clearly it is linear map. The dimension of $Xa$ is $3\times1$. But then we can't choose $[T]$ such that dimension of $[T]X$ is equal to $Xa$.

$[T]$ is matrix corresponding to given linear map.

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    What is the domain and codomain of $T$? – 2017-02-20
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    @K.Miller Actually in the problem they even asked to find domain and codomain of $T$. – 2017-02-20
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    Ok, so let's assume that the domain is the set of $3 \times 3$ matrices over $\mathbb{R}$ and the codomain is $\mathbb{R}^3$. – 2017-02-20

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Let $T : \mathbb{R}^{3\times 3} \to \mathbb{R}^3$ be a linear map defined by $$ T(X) = Xa $$ for some vector $a\in\mathbb{R}^3$. If $a$ is the zero vector then clearly the nullspace of $T$ is equal to $\mathbb{R}^{3\times 3}$ and the range of $T$ is $\{0\}$. Now suppose $a \neq 0$. Let $X_{ij} = e_ie_j^T$ for $i,j = 1,2,3$ denote the standard basis vectors for $\mathbb{R}^{3\times 3}$. Then \begin{align} [T] &= [T(X_{11})~T(X_{21})~T(X_{31})~T(X_{12})~T(X_{22})~T(X_{32})~T(X_{13})~T(X_{23})~T(X_{33})]\\[1mm] &= \begin{bmatrix} a_1 & 0 & 0 & a_2 & 0 & 0 & a_3 & 0 & 0\\ 0 & a_1 & 0 & 0 & a_2 & 0 & 0 & a_3 & 0\\ 0 & 0 & a_1 & 0 & 0 & a_2 & 0 & 0 & a_3 \end{bmatrix} \end{align} Without restricting generality suppose $a_1 \neq 0$. Then the RREF of $[T]$ is $$ \begin{bmatrix} 1 & 0 & 0 & a_2/a_1 & 0 & 0 & a_3/a_1 & 0 & 0\\ 0 & 1 & 0 & 0 & a_2/a_1 & 0 & 0 & a_3/a_1 & 0\\ 0 & 0 & 1 & 0 & 0 & a_2/a_1 & 0 & 0 & a_3/a_1 \end{bmatrix} $$ Thus, the dimension of the nullspace of $[T]$ is $6$. It follows that $[T][X] = [0]$ if and only if $$ X = \begin{bmatrix} -\frac{a_2}{a_1}x_{12} -\frac{a_3}{a_1}x_{13} & x_{12} & x_{13}\\ -\frac{a_2}{a_1}x_{22} -\frac{a_3}{a_1}x_{23} & x_{22} & x_{23}\\ -\frac{a_2}{a_1}x_{32} -\frac{a_3}{a_1}x_{33} & x_{32} & x_{33} \end{bmatrix} $$