Step 1: Find the derivative of f(x)

To determine when the function $f(x)$ is decreasing, we need to find its derivative $f'(x)$ and analyze when $f'(x) \le 0$ for all $x$ in R.

Step 2: Calculate f'(x)

Given $f(x) = \sqrt{3}\sin x - \cos x - 2ax + 6$, we differentiate with respect to $x$:

$$f'(x) = \sqrt{3}\cos x + \sin x - 2a$$

Step 3: Apply the decreasing condition

For $f(x)$ to be decreasing in R, we must have $f'(x) \le 0$ for all $x$ in R. Therefore,

$$\sqrt{3}\cos x + \sin x - 2a \le 0$$

$$2a \ge \sqrt{3}\cos x + \sin x$$

Step 4: Find the maximum value of the trigonometric expression

We need to find the maximum value of $\sqrt{3}\cos x + \sin x$. We can rewrite this expression in the form $R\cos(x - \alpha)$, where $R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$.

So, $\sqrt{3}\cos x + \sin x = 2\cos(x - \alpha)$ for some angle $\alpha$.

The maximum value of $2\cos(x - \alpha)$ is 2, since the maximum value of $\cos(x - \alpha)$ is 1.

Step 5: Determine the condition for 'a'

Thus, we have $2a \ge 2$, which implies $a \ge 1$.