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Here's Theorem 4.3-2 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $X$ be a normed space, let $Z$ be a subspace of $X$, and let $f$ be a bounded linear functional defined on $Z$. Then there exists a bounded linear functional $\tilde{f}$ defined on $X$ such that $$\tilde{f}(x) = f(x) \ \mbox{ for all } \ x \in Z, \ \mbox{ and } \ \Vert \tilde{f} \Vert_X = \Vert f \Vert_Z,$$ where $$\Vert f \Vert_Z \colon= \sup \{ \ \frac{\vert f(z) \vert }{\Vert z \Vert} \ \colon \ z \in Z, \ z \neq 0 \ \} \ \mbox{ if } \ Z \neq \{ 0 \}; \ \mbox{ otherwise } \ \Vert f \Vert_Z = 0.$$ And, $$\Vert \tilde{f} \Vert_X \colon= \sup \{ \ \frac{ \vert \tilde{f}(x) \vert }{ \Vert x \Vert } \ \colon \ x \in X, \ x \neq 0 \ \}.$$

Now let $X \colon= \mathbb{R}^3$ with the Euclidean norm, let $a \colon= (\alpha_1, \alpha_2, 0) \in X$, and let $$Z \colon= \{ \ (\xi_1, \xi_2, \xi_3) \in \mathbb{R}^3 \ \colon \ \xi_3 = 0 \ \}.$$ Let $f$ be defined on $Z$ by $$f(z) \colon= \alpha_1 \xi_1 + \alpha_2 \xi_2 \ \mbox{ for all } \ z \colon= (\xi_1, \xi_2, 0) \in Z.$$ Then what are all the possible linear extensions $\tilde{f}$ of $f$ as gauranteed by Theorem 4.3-2 in Kreyszig?

Here, $$\Vert f \Vert = \Vert a \Vert = \sqrt{ \alpha_1^2 + \alpha_2^2}.$$

Of course, one possible extension $\tilde{f}$ is given by $$\tilde{f}(x) \colon= \alpha_1 \xi_1 + \alpha_2 \xi_2 \ \mbox{ for all } \ x \colon= (\xi_1, \xi_2, \xi_3) \in \mathbb{R}^3. $$

Am I right?

If so, then what are other such extensions $\tilde{f}$, if any?

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Any bounded linear functional defined on a subspace of a Hilbert space admits a unique norm preserving extension.

The proof is given here.

More on Banach spaces with unique extension property for functionals you can find in this discussion

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Your analysis is correct.

Also your $\tilde f$ is the only such extension. By the Riesz representation theorem, every bounded linear functional g on $\mathbb{R}^3$ (with Euclidean norm) is of the form $g(x) = \left<x, y \right>$ for some $y \in \mathbb{R}^3$. If $g$ is defined by $y = (\alpha_1,\alpha_2,\alpha_3)$ for some $\alpha_3 \neq 0$, then $\lVert g\rVert = \sqrt{\alpha_1^2 + \alpha_2^2 +\alpha_3^2 } > \sqrt{\alpha_1^2 + \alpha_2^2 } = \lVert f \rVert $.

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