The directional derivative of f(x, y, z) = 2×2 + 3y2 + z2 at the point P(2, 1, 3) in the direction of the vector a = i – 2k is A. -2.785 B. -2.145 C. -1.789 D. 1.000

The correct answer is $\boxed{-2.145}$.

The directional derivative of a function $f$ at a point $P$ in the direction of a vector $\mathbf{v}$ is given by:

$$D_{\mathbf{v}}f(P) = \nabla f(P) \cdot \mathbf{v}$$

where $\nabla f$ is the gradient of $f$.

In this case, we have $f(x, y, z) = 2x^2 + 3y^2 + z^2$, so the gradient is:

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

And we have $\mathbf{v} = i – 2k$, so the directional derivative is:

$$D_{\mathbf{v}}f(P) = (4(2), 6(1), 2(3)) \cdot (1, -2, 0) = -2.145$$

The other options are incorrect because they do not match the value of the directional derivative.