The condition $$\vec a+\vec b+\vec c=\vec 0$$ means that if we place the three vectors head-to-tail they form the three consecutive sides of a triangle.
Let the side-lengths of that triangle be
$$|\vec a|=x,\;|\vec b|=y,\;|\vec c|=z,$$
so that
$$x+y+z=12.\qquad -(1)$$
Label the interior angles of the triangle by $$A,B,C$$ such that angle $$A$$ lies between the sides of lengths $$y$$ and $$z$$, etc.
Because $$\vec a$$ and $$\vec b$$ meet at angle $$B$$ we have
$$|\vec a\times\vec b| = xy\sin B,$$
and cyclically
$$|\vec b\times\vec c| = yz\sin C,\qquad |\vec c\times\vec a| = zx\sin A.$$
Hence
$$P=|\vec a\times\vec b|+|\vec b\times\vec c|+|\vec c\times\vec a|
=xy\sin B+yz\sin C+zx\sin A. \qquad -(2)$$
Apply the sine rule $$x/\sin A=y/\sin B=z/\sin C=2R,$$ where $$R$$ is the circum-radius of the triangle.
From it we get
$$\sin A=\dfrac{x}{2R},\;\sin B=\dfrac{y}{2R},\;\sin C=\dfrac{z}{2R}.$$
Substituting these values in (2) gives
$$P=\frac1{2R}\left(xy^2+yz^2+zx^2\right). \qquad -(3)$$
For a fixed perimeter the equilateral triangle (i) minimises the circum-radius and (ii) maximises the symmetric quadratic $$xy^2+yz^2+zx^2$$.
Therefore $$P$$ attains its maximum when the triangle is equilateral.
Case of an equilateral triangle
Put $$x=y=z=4$$ (because $$3\cdot4=12$$).
Its circum-radius is $$R=\dfrac{x}{\sqrt3}=\dfrac4{\sqrt3},$$ so $$2R=\dfrac8{\sqrt3}.$$
The numerator in (3) becomes
$$xy^2+yz^2+zx^2=4\cdot4^2+4\cdot4^2+4\cdot4^2=3\cdot4\cdot16=192.$$
Hence
$$P_{\max}= \frac{192}{8/\sqrt3}=24\sqrt3.$$
This value is attainable (take three 120° vectors of length 4), and every other admissible triangle gives a smaller value by the reasoning above. Therefore
Maximum value of $$P$$ = $$24\sqrt3$$.
Option A which is: $$24\sqrt3$$
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