SOLUTION:-
\[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}\]
As per the inquiry,
A line section\[~AB\] in the proportion \[4:7\]
Thus, \[A:B\text{ }=\text{ }4:7\]
Presently,
Draw a beam \[AX\]making an acute angle \[BAX\]
Least number of focuses situated at equivalent distances on the beam,
Hatchet \[=\text{ }A+B\text{ }=\text{ }4+7=\text{ }11\]
\[{{A}_{1}},\text{ }{{A}_{2}},\text{ }{{A}_{3}},\text{ }\ldots \ldots \ldots .\]
are situated at equivalent distances on the beam \[AX.\]
Point \[B\]is joined to the last point is
\[~{{A}_{11}}.\]