In $\mathbb{R}^3$, consider a line segment $[AB]$ and a line $l$ containing $C$. Then we have angle $\angle ACB$ at $C$. We want to find an angle smaller than it.
Assume that ${\rm area}\ \Delta ABC'$ is smallest s.t. $C'\in l$, where $\Delta xyz$ is a triangle with vertices $x,\ y,\ z$ and ${\rm area}$ is a area function. Then
$\angle AC'B \leq \angle ACB$.
Is this right ? Thank you in advance.
Case 1 : $ {\rm dim}\ {\rm conv}\ [AB]\cup l =3 $
Note that there is a plane $P$ containing $\Delta ABC'$ s.t. $P$ is perpendicular to $l$.
Let $$t:=|C-C'|,\ f(t)=\angle ACB,\ |B-C|=a,\ |A-C|=b,\ |A-B|=c$$ By Cosine Law $$ a^2+b^2+2t^2 -2\sqrt{a^2+t^2}\sqrt{b^2+t^2}\cos\ f=c^2 $$
So \begin{align*} & 0<f<\pi,\ -\sin\ ff' \\& \\&= \frac{d}{dt} \frac{c^2-a^2-b^2-2t^2}{-2 \sqrt{a^2+t^2}\sqrt{b^2+t^2} } \\&\\&= \frac{t \{ -(a^2-b^2)^2 + (a^2+b^2+2t^2)c^2 \} }{4 ( \sqrt{a^2+t^2}\sqrt{b^2+t^2})^2} \end{align*}
Hence if $a=b$, then $f'(t)<0$ for all $t$.
By considering $a=b+c$ we have $f'(t) >0$ for $0<t<\epsilon$ if $a$ is close to $b+c$.
Case 2 : $ {\rm dim}\ {\rm conv}\ [AB]\cup l =2 $
If $l'$ is a line containing $[AB]$ and if $l'$ intersects $l$, then $\angle AC'B=0$
If $l'$ is parallel to $l$, then $C'$ is not uniquely determined.
Consider $A=(0,0),\ B=(\varepsilon, 0),\ C=(x,1)$ Let $f(x)=\angle ACB$. Then by Cosine Law $$ x^2+1+(x-\varepsilon)^2 +1 -2\sqrt{x^2+1}\sqrt{ (x- \varepsilon)^2 +1} \cos\ f=\varepsilon^2 $$
$$ -\sin\ ff' = \frac{ -\varepsilon^2(\varepsilon-2x) }{ (\sqrt{ x^2+1}\sqrt{ (x- \varepsilon)^2 +1} )^3 } $$
Hence $f'(x)<0$ for $x> \frac{\varepsilon }{2}$.