EDIT: Sorry, I realised after a moment that I was thinking 'realistically'; this map is not good, since $\ell' = \ell$ when $\ell = \mathbb C\cdot(1, i, 1, i)$. I tried to delete this answer but could not, since it has been accepted.

The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot(b, -a, d, -c)$ when $\ell = \mathbb C\cdot(a, b, c, d)$, is holomorphic, and satisfies your first condition.

The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot\overline{(b, -a, d, -c)}$ when $\ell = \mathbb C\cdot(a, b, c, d)$, satisfies your first condition.

(I originally had a version without the complex conjugation, which doesn't work because $\ell' = \ell$ when $\ell = \mathbb C\cdot(1, i, 1, i)$. Fortunately @AliTaghavi pointed out how to fix it. The candidate without the conjugation would have been holomorphic, hence contradicted @abx's answer.)

The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot(b, -a, d, -c)$ when $\ell = \mathbb C\cdot(a, b, c, d)$, is holomorphic, and satisfies your first condition.

EDIT: Sorry, I realised after a moment that I was thinking 'realistically'; this map is not good, since $\ell' = \ell$ when $\ell = \mathbb C\cdot(1, i, 1, i)$. I tried to delete this answer but could not, since it has been accepted.

The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot(b, -a, d, -c)$ when $\ell = \mathbb C\cdot(a, b, c, d)$, is holomorphic, and satisfies your first condition.