For a scalar function f(x, y, z) = x2 + 3y2 + 2z2, the gradient at the point P(1, 2, -1) is A. $$2\overrightarrow {\text{i}} + 6\overrightarrow {\text{j}} + 4\overrightarrow {\text{k}} $$ B. $$2\overrightarrow {\text{i}} + 12\overrightarrow {\text{j}} – 4\overrightarrow {\text{k}} $$ C. $$2\overrightarrow {\text{i}} + 12\overrightarrow {\text{j}} + 4\overrightarrow {\text{k}} $$ D. $$\sqrt {56} $$

The correct answer is $\boxed{\text{B}}$.

The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field, and its magnitude is equal to the magnitude of the rate of increase.

The gradient of a scalar field $f$ is given by the formula:

$$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$

In this case, we have $f(x, y, z) = x^2 + 3y^2 + 2z^2$. Therefore, the gradient of $f$ is given by:

$$\nabla f = \left(2x, 6y, 4z\right)$$

Evaluating the gradient at the point $P(1, 2, -1)$ gives:

$$\nabla f(1, 2, -1) = (2, 12, -4) = \boxed{2\overrightarrow {\text{i}} + 12\overrightarrow {\text{j}} – 4\overrightarrow {\text{k}}}$$

Option A is incorrect because it does not take into account the $z$-coordinate of the point $P(1, 2, -1)$. Option C is incorrect because it does not take into account the $y$-coordinate of the point $P(1, 2, -1)$. Option D is incorrect because it is the magnitude of the gradient, not the gradient itself.