The given parabola is \[{{x}^{2}}~=\text{ }12y.\]
On contrasting this condition and\[{{x}^{2}}~=\text{ }4ay\], we get,
\[\begin{array}{*{35}{l}}
4a\text{ }=\text{ }12 \\
a\text{ }=\text{ }12/4 \\
=\text{ }3 \\
\end{array}\]
The directions of foci are \[S\left( 0,a \right)\text{ }=\text{ }S\left( 0,3 \right).\]
Presently let AB be the latus rectum of the given parabola.
The given parabola can be generally drawn as
\[\begin{array}{*{35}{l}}
At\text{ }y\text{ }=\text{ }3,\text{ }{{x}^{2}}~=\text{ }12\left( 3 \right) \\
{{x}^{2}}~=\text{ }36 \\
x\text{ }=~\pm 6 \\
\end{array}\]
In this way, the directions of \[A\text{ }are\text{ }\left( -\text{ }6,\text{ }3 \right),\]while the directions of \[B\text{ }are\text{ }\left( 6,\text{ }3 \right)\]
Then, at that point, the vertices of \[\Delta OAB\text{ }are\text{ }O\left( 0,0 \right),\text{ }A\text{ }\left( -\text{ }6,3 \right)\text{ }and\text{ }B\left( 6,3 \right).\]
By utilizing the equation,
\[\begin{array}{*{35}{l}}
Area\text{ }of\text{ }\Delta OAB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left[ 0\left( 3-3 \right)\text{ }+\text{ }\left( -6 \right)\left( 3-0 \right)\text{ }+\text{ }6\left( 0-3 \right) \right]\text{ }uni{{t}^{2}} \\
=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left[ \left( -6 \right)\text{ }\left( 3 \right)\text{ }+\text{ }6\text{ }\left( -3 \right) \right]\text{ }uni{{t}^{2}} \\
=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left[ -18-18 \right]\text{ }uni{{t}^{2}} \\
=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left[ -36 \right]\text{ }uni{{t}^{2}} \\
=\text{ }18\text{ }uni{{t}^{2}} \\
\therefore \text{ }Area\text{ }of\text{ }\Delta OAB\text{ }is\text{ }18\text{ }uni{{t}^{2}} \\
\end{array}\]