Answer
$26.1 N$
Work Step by Step
Let the two forces in the plane be
\[
\vec{F}_1 = 25\ \text{N},\quad \vec{F}_2 = 12\ \text{N},
\]
with an angle of \(100^\circ\) between them. Their resultant \(\vec{R}\) in the plane is given by
\[
R = \sqrt{25^2 + 12^2 + 2\cdot25\cdot12\cos(100^\circ)}.
\]
Calculating each term:
\[
25^2 = 625,\quad 12^2 = 144,\quad 2\cdot25\cdot12 = 600.
\]
Since \(\cos(100^\circ) \approx -0.17365\),
\[
R \approx \sqrt{625 + 144 + 600(-0.17365)} = \sqrt{769 - 104.19} \approx \sqrt{664.81} \approx 25.79\ \text{N}.
\]
The third force, \(\vec{F}_3\), has a magnitude of \(4\ \text{N}\) and is perpendicular to the plane of \(\vec{F}_1\) and \(\vec{F}_2\). Thus, the net force vector \(\vec{F}_{\text{net}}\) is the vector sum of \(\vec{R}\) (in the plane) and \(\vec{F}_3\) (perpendicular to the plane). Since these components are perpendicular, the magnitude of the net force is
\[
F_{\text{net}} = \sqrt{R^2 + 4^2} \approx \sqrt{(25.79)^2 + 16} \approx \sqrt{664.81 + 16} \approx \sqrt{680.81} \approx 26.11\ \text{N}.
\]
A force of equal magnitude but opposite direction is required to counterbalance these forces. Hence, the magnitude of the counterbalancing force is
\[
\boxed{26.1\ \text{N}}.
\]
