**Finding the Directional Derivative**

To find the directional derivative of the given function, varphi(x, y, z) = x^2 * y * z + 4x * z^2, at the point (1, -2, -1) in the direction 2i - j - 2k, we need to calculate the dot product of the gradient of the function with the given direction vector.

The directional derivative can be calculated using the formula:

D_v(f) = ∇f · v

where D_v(f) is the directional derivative of f in the direction of vector v, ∇f is the gradient of f, and · represents the dot product.

**Calculating the Gradient**

The gradient of a scalar function f(x, y, z) is given by:

∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k

For our given function, varphi(x, y, z) = x^2 * y * z + 4x * z^2, let's calculate the partial derivatives:

∂varphi/∂x = 2xy * z + 4z^2
∂varphi/∂y = x^2 * z
∂varphi/∂z = x^2 * y + 8xz

Therefore, the gradient of varphi is:

∇varphi = (2xy * z + 4z^2)i + (x^2 * z)j + (x^2 * y + 8xz)k

**Calculating the Directional Derivative**

Now that we have the gradient ∇varphi and the direction vector v = 2i - j - 2k, we can calculate the directional derivative D_v(varphi) using the dot product:

D_v(varphi) = ∇varphi · v

Substituting the values:

D_v(varphi) = (2xy * z + 4z^2)(2) + (x^2 * z)(-1) + (x^2 * y + 8xz)(-2)

Simplifying further:

D_v(varphi) = 4xy * z + 8z^2 - x^2 * z - 2x^2 * y - 16xz

**Final Answer**

Therefore, the directional derivative of varphi(x, y, z) = x^2 * y * z + 4x * z^2 at the point (1, -2, -1) in the direction 2i - j - 2k is given by the expression:

D_v(varphi) = 4xy * z + 8z^2 - x^2 * z - 2x^2 * y - 16xz

This represents the rate of change of the function varphi in the direction of the given vector at the specified point.