Is my attempt at the following proof correct?
PRELIMINARY NOTATION AND PROOFS
$\mathcal{L}(\mathbf{R^3})$ denotes the set of Linear operator on $\mathbf{R^3}$.
$5.10$ is the result that eigenvectors of a Linear operator $T$ corresponding to distinct eigenvalues form a linearly independent list.
Theorem. Given that $T\in\mathcal{L}(\mathbf{R^3})$ such that $-4,5$ and $\sqrt{7}$ are eigenvalues of $T$. There exists a $x\in\mathbf{R^3}$ such that $Tx-9x = (-4,5,\sqrt{7})$.
Proof. Let $v_1,v_2,v_3$ be the eigenvectors corresponding to $-4,5$ and $\sqrt{7}$ respectively. Since $-4,5,\sqrt{7}$ are distinct it follows by theorem $\textbf{5.10}$ that the vectors $v_1,v_2,v_3$ are linearly independent and thus a basis for $\mathbf{R^3}$ but then so is the list $v_1+\overset{9}w_1,v_1+\overset{9}w_2,v_1+\overset{9}w_3$ where $\overset{9}w_i$ denotes a column vector of length $3$ with the $i^{th}$ row equal to $9$ and all others $0$.
With $v_1+\overset{9}w_1,v_1+\overset{9}w_2,v_1+\overset{9}w_3$ as a basis for $V$ it follows that the matrix of $T$ i.e. $\mathcal{M}(T)$ is as follows $$ \mathcal{M}(T) = \begin{pmatrix} 5&0&0\\ 0&14&0\\ 0&0&\sqrt{7}+9 \end{pmatrix} $$ and as a consequence that of the transformation $T-9I$ i.e. $\mathcal{M}(T-9I)$ is $$ \mathcal{M}(T-9I) = \begin{pmatrix} -4&0&0\\ 0&5&0\\ 0&0&\sqrt{7} \end{pmatrix} $$ Thus the vector $x = (1,1,1)$ yields the required result.
I think that this is only true if $v_k+w_k=e_k$ for $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. But you can write $\mathcal M(T)$ as a matrix after a change of basis, so your proof will remain correct, that is, there is a linear map $B\in\mathcal L(\Bbb R^3)$ such that
$$\mathcal M(BTB^{-1})=\begin{pmatrix} 5&0&0\\ 0&14&0\\ 0&0&\sqrt{7}+9 \end{pmatrix}$$
Thus
$$\mathcal M(B(T-9I)B^{-1})=\begin{pmatrix} -4&0&0\\ 0&5&0\\ 0&0&\sqrt{7} \end{pmatrix}$$
where $B$ is defined by $B(v_k+w_k)=e_k$.
A shorter proof becomes from noticing that $T-9I$ is bijective, then such vector must exists.