Explanation:

Given, f = x^2yz + xz^2 and the point of interest is (1, -2, -1).

The directional derivative of f in the direction of a unit vector u = 2i - j - 2k is given by the dot product of the gradient of f at the point (1, -2, -1) and the unit vector u.

Step 1: Find the gradient of f at the point (1, -2, -1)

∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k
= (2xyz + z^2)i + (x^2z)j + (x^2y + 2xz)k

∇f at (1, -2, -1) = (2(-2)(-1) + (-1)^2)i + (1^2(-1))j + (1^2(-2) + 2(1)(-1))k
= 4i - 1j - 4k

Step 2: Find the dot product of the gradient of f and the unit vector u

u = 2i - j - 2k
|u| = √(2^2 + (-1)^2 + (-2)^2) = √9 = 3
Unit vector u = (2/3)i - (1/3)j - (2/3)k

∇f at (1, -2, -1) . u = (4i - j - 4k) . ((2/3)i - (1/3)j - (2/3)k)
= (8/3) - (1/3) - (8/3)
= -1

Step 3: Interpretation

The directional derivative of f in the direction of the unit vector u = 2i - j - 2k at the point (1, -2, -1) is -1. This means that the rate of change of f in the direction of the vector u at the point (1, -2, -1) is -1.