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This is not an answer, just a remark. I find your question interesting even for small values of $n=|V|$. For example, if $n=4$, then choosing $V$ as the vertices of a regular tetrahedron, the $\binom{4}{2}=6$ planes $V_P$ determine six normal lines that define a cuboctahedron:

The angles between these lines/planes is either $90^\circ$ or about $63.6^\circ$ (or $116.4^\circ$) $60^\circ$ (or $120^\circ$). Now, the optimal packing of six lines is known (Conway, Hardin, Sloane) to be the six diameters of the icosahedron. The minimum angle determined by those diameters is a bit larger, if I've calculated correctly: $63.4^\circ$.

So: Is there an arrangement of four vectors $V$ that yields this optimal line packing? Answer: No, $60^\circ$ is the optimal for $n=4$. See Henry Cohn's argument in the comments.

Here is Edmund's suggestion for $n=6$: $V$ is given by half the vertices of an icosahedron (blue), which generate 15 normal lines (red) passing through the midpoints of the icosahedron's 30 edges, with the normal lines separated by $49.7^\circ$.

The tips of the normal lines form the vertices of an icosidodecahedron.

4 added 62 characters in body

This is not an answer, just a remark. I find your question interesting even for small values of $n=|V|$. For example, if $n=4$, then choosing $V$ as the vertices of a regular tetrahedron, the $\binom{4}{2}=6$ planes $V_P$ determine six normal lines that define a cuboctahedron:

The angles between these lines/planes is either $90^\circ$ or about $63.6^\circ$ (or $116.4^\circ$) $60^\circ$ (or $120^\circ$). Now, the optimal packing of six lines is known (Conway, Hardin, Sloane) to be the six diameters of the icosahedron. The minimum angle determined by those diameters is a bit larger, if I've calculated correctly: $63.4^\circ$.

So: Is there an arrangement of four vectors $V$ that yields this optimal line packing? Answer: No, $60^\circ$ is the optimal for $n=4$. See Henry Cohn's argument in the comments.

Here is Edmund's suggestion for $n=6$: Half $V$ is given by half the vertices of an icosahedron (blue) blue), which generate 15 normal vectors lines (red) passing through the midpoints of the icosahedron's 30 edges, with the normal lines separated by $49.7^\circ$.

The tips of the normal lines form the vertices of an icosidodecahedron.

3 added 594 characters in body

This is not an answer, just a remark. I find your question interesting even for small values of $n=|V|$. For example, if $n=4$, then choosing $V$ as the vertices of a regular tetrahedron, the $\binom{4}{2}=6$ planes $V_P$ determine six normal lines that define a cuboctahedron:

The angles between these lines/planes is either $90^\circ$ or about $63.6^\circ$ (or $116.4^\circ$) $60^\circ$ (or $120^\circ$). Now, the optimal packing of six lines is known (Conway, Hardin, Sloane) to be the six diameters of the icosahedron. The minimum angle determined by those diameters is a bit larger, if I've calculated correctly: $63.4^\circ$.

So: Is there an arrangement of four vectors $V$ that yields this optimal line packing? Answer: No, $60^\circ$ is the optimal for $n=4$. See Henry Cohn's argument in the comments.

Here is Edmund's suggestion for $n=6$: Half the vertices of an icosahedron (blue) generate 15 normal vectors passing through the midpoints of the icosahedron's 30 edges, separated by $49.7^\circ$.

The normal lines form the vertices of an icosidodecahedron.

2 Corrected as per Henry.; deleted 10 characters in body
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