Consider a cube of unit edge length and sides parallel to co-ordinate axes, with its centroid at the point (1, 2, 3). The surface integral ${\int }_A\overrightarrow{F}.d\overrightarrow{A}$ of a vector field $\overrightarrow{F}=3x\hat{i}+5y\hat{j}+6z\hat{k}$ over the entire surface A of the cube is ______.

The correct option is 14.

To find the surface integral of the vector field over the entire closed surface of the cube, we can apply Gauss's Divergence Theorem. Gauss's Divergence Theorem states that the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence of that vector field over the volume enclosed by the surface.

Mathematically, the theorem is expressed as:
${\int }_A\overrightarrow{F}\cdot d\overrightarrow{A}={\int }_V\left(\nabla \cdot \overrightarrow{F}\right)dV$
where $V$ is the volume of the cube and $A$ is its boundary surface.

First, let's calculate the divergence of the given vector field $\overrightarrow{F}=3x\hat{i}+5y\hat{j}+6z\hat{k}$:

$\nabla \cdot \overrightarrow{F}=\frac{\partial }{\partial x}\left(3x\right)+\frac{\partial }{\partial y}\left(5y\right)+\frac{\partial }{\partial z}\left(6z\right)$

Evaluating the partial derivatives, we get:
$\nabla \cdot \overrightarrow{F}=3+5+6=14$

Now, we substitute this constant divergence back into the volume integral:
${\int }_V\left(\nabla \cdot \overrightarrow{F}\right)dV={\int }_{V}14dV=14{\int }_{V}dV$
where ${\int }_{V}dV$ represents the volume $V$ of the cube.

The problem states that the cube has a unit edge length (side length $L=1$). The volume of a cube is given by $V=L^3$, which gives:
$V=1^3=1$

Thus, the value of the surface integral is:
$14\cdot V=14\cdot 1=14$