Finding image and nucleus of $ f: x \in \mathbb{R}^3 \mapsto (aa^T +bb^T)x \in \mathbb{R}^3 $ I got a problem.

Let $a, b ∈ \mathbb{R}^3$ be two linearly independent nonzero vectors
in a three-dimensional real space, and let $c ∈ \mathbb{R}^3$ be a
nonzero vector orthogonal to $a, b$. Find the image and nucleus of the
linear map:
$$ f: x \in \mathbb{R}^3 \mapsto (aa^T +bb^T)x \in \mathbb{R}^3 $$

I tried to solve this, but what came through my head is:
$$
\operatorname{Im}{f} = \left\{\left(\begin{array}{l}
\left(a_{1}^{2}+b_{1}^{2}\right) x_{1}+\left(a_{1} a_{2}+b_{1} b_{2}\right) x_{2}+\left(a_{1} a_{3}+b_{1} b_{3}\right) x_{3} \\
\left(a_{1} a_{2}+b_{1} b_{2}\right) x_{1}+\left(a_{2}^{2}+b_{2}^{2}\right) x_{2}+\left(a_{2} a_{3}+b_{2} b_{3}\right) x_{3} \\
\left(a_{1} a_{3}+b_{1} b_{3}\right) x_{1}+\left(a_{2} a_{3}+b_{2} b_{3}\right) x_{2}+\left(a_{3}^{2}+b_{3}^{2}\right) x_{3}
\end{array}\right) \, \middle| \, x_1, x_2, x_3 \in \mathbb{R}\right\}
$$
I also do not understand what this question would imply.
Could you give me some hints to solve this?

Observe that $f(a)=aa^Ta+bb^Ta=\|a\|^2\cdot a$, $f(b)=aa^Tb+bb^Tb=\|b\|^2\cdot b $ and $f(c)=aa^T c + bb^T c = 0$. Therefore the kernel of $f$ is $c$ and the image of $f$ is $\mathrm{span}\{ a,b \}$.
