Riemannian metric for euclidean geometry

I am a beginner to Riemannian geometry. Following is my question.

In the Euclidean space, say $\mathbb{R}^3$, let us consider a plane, for simplicity, say one passing through the origin, $\mathbb{P}=\{\textbf{x}=(x_1,x_2,x_3):ax_1+bx_2+cx_3=0\}$ and consider the origin $\textbf{0}=(0,0,0)\in \mathbb{P}$.

Now, let us consider the set of all points ($\mathbb{L}$) in $\mathbb{R}^3$ whose best approximation with respect to $\ell_2$-distance on $\mathbb{P}$ is $\textbf{0}$, i.e., $$\{\textbf{y}:\arg \min_{\textbf{x}\in \mathbb{P}}\|\textbf{x}-\textbf{y}\|_2=\textbf{0}\}$$ which is precisely the line $$\mathbb{L}:~~~~~~~\frac{x_1}{a}=\frac{x_2}{b}=\frac{x_3}{c}$$

This line $\mathbb{L}$ can also be characterised by $\mathbb{L}=\{\textbf{y}:\langle \textbf{x},\textbf{y}\rangle=0 \text{ for all } \textbf{x}\in \mathbb{P}\}$.

My question is, how to define a Riemannian metric $g$ with components $g_{i,j}$ and a connection $\nabla$ with connection coefficients $\Gamma_{i,j}^k$ so that $\mathbb{L}$ is a $\nabla$-geodesic and prove that it intersects the plane $\mathbb{P}$ at the single point $\textbf{0}$?

Thank you.

-
The Riemannian metric of what? – Mariano Suárez-Alvarez Feb 16 '13 at 6:31
Do you wish to equip $\mathbb{L}$ with a Riemannian metric? – Haskell Curry Feb 16 '13 at 6:33
For $\mathbb{R}^3$. – Kumara Feb 16 '13 at 6:33
@MarianoSuárez-Alvarez and Haskell Curry: I haven't understood Riemannian geometry yet. This is question is towards understanding it better. I just wanted to define a Riemannian metric on $\mathbb{R}^3$ and prove that result in that framework. Am I making sense? – Kumara Feb 16 '13 at 6:38
Let $e_1, e_2, e_3$ be constant, orthonormal basis vectors for $\mathbb R^3$. The metric $g$ is just $g(e_i, e_j) = \delta_{ij} = \langle e_i, e_j \rangle$. – Muphrid Feb 16 '13 at 7:10