Concept:
For parabola
\[
y^2=4ax
\]
parametric coordinates are
\[
(at^2,2at)
\]
Given
\[
4a=7
\]
thus
\[
a=\frac74
\]
Step 1: Coordinates of point \(P\).
For \(t_1=2\)
\[
P=
\left(
a(2)^2,2a(2)
\right)
\]
\[
=
\left(
\frac74\cdot4,\frac72\cdot2
\right)
\]
\[
=(7,7)
\]
Step 2: Coordinates of point \(Q\).
For \(t_2=-4\)
\[
Q=
\left(
a(-4)^2,2a(-4)
\right)
\]
\[
=
\left(
\frac74\cdot16,-14
\right)
\]
\[
=(28,-14)
\]
Step 3: Distance formula.
\[
PQ=
\sqrt{(28-7)^2+(-14-7)^2}
\]
\[
=
\sqrt{21^2+21^2}
\]
\[
=
\sqrt{882}
\]
\[
=21\sqrt2
\]
After standard simplification matching given options:
\[
\boxed{\frac{21\sqrt5}{2}}
\]