Find $C\in\mathbb R$ such that $(2,3,5)$ be in $\text{Im}(F)$ Let $$\color{red}{\underbrace{F:\mathbb R^3\to\mathbb R^3\;|}_{\text{Added by me}}}\; F(x,y,z)=(x+Cz,\;x+y+2z,\;x+Cy)$$ a linear transformation. Find $C\!\!\!\!\!\color{red}{\underbrace{\in\mathbb R}_{\text{Added by me}}}\!\!\!\!\!\!$ such that $(2,3,5)$ be in $\text{Im}(F)$.

I don't know hot to interpet the statement. I think I need to solve the equation $$F(x,y,z)=(2,3,5)\qquad\text{i.e.}\qquad(x+Cz,\;x+y+2z,\;x+Cy)=(2,3,5),$$ but this is a system equations of $3\times 4$. Is correct my reasoning? If so, how can it solve to find $C$?
Thank you!
 A: $$(x+Cz, x+y+2z, x+Cy)=(2,3,5)$$
$$\implies x+Cz=2;~~~ x+y+2z=3;~~~~ x+Cy=5$$
$$\begin{bmatrix}1&0&C\\1&1&2\\1&C&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}$$
applying row transformation, $R_2\leftarrow R_2-R_1$, $R_3\leftarrow R_3-R_1$
$$\begin{bmatrix}1&0&C\\0&1&2-C\\0&C&-C\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$$
$R_3\leftarrow R_3-C R_2$
$$\begin{bmatrix}1&0&C\\0&1&2-C\\0&0&C^2-3C\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3-C\end{bmatrix}$$
For unique solution $(x,y,z)$, we take $C$ different from $3$.
NOTE:- for $C=0$, rank of coefficient matrix is not equal to rank of augmented matrix. Hence, for $C=0$ solution does not exist.
Take $C=1$
$$\begin{bmatrix}1&0&1\\0&1&1\\0&0&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\2\end{bmatrix}$$
$R_3\leftarrow \frac{-1}{2}R_3$ followed by $R_2\leftarrow R_2-R_3$, $R_1\leftarrow R_1-R_3$
$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\2\\-1\end{bmatrix}$$
$$x=3;~~y=2;~~z=-1.$$
$$(x+Cz, x+y+2z, x+Cy)=(3+1(-1),3+2+2(-1),3+1(2))=(2,3,5)$$
