A function f(x, y, z) = x^2 + y^2 + z^2 at the point (1, 2, 3) in the direction of vector v = <2, -1,="" 3="">.

To find the directional derivative, we need to take the dot product of the gradient of the function with the unit vector in the direction of vector v.

The gradient of f(x, y, z) is given by ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).

Taking partial derivatives with respect to x, y, and z, we have:
∂f/∂x = 2x
∂f/∂y = 2y
∂f/∂z = 2z

So, the gradient of f(x, y, z) is ∇f = (2x, 2y, 2z).

To find the unit vector in the direction of vector v, we need to divide vector v by its magnitude:

|v| = sqrt(2^2 + (-1)^2 + 3^2) = sqrt(4 + 1 + 9) = sqrt(14)

The unit vector in the direction of vector v is v/|v| = <2 qrt(14),="" -1/sqrt(14),="" 3/sqrt(14)="">.

Now, we can calculate the directional derivative by taking the dot product of ∇f and the unit vector:

∇f · (v/|v|) = (2x, 2y, 2z) · (2/sqrt(14), -1/sqrt(14), 3/sqrt(14))
= (4x/sqrt(14)) + (-2y/sqrt(14)) + (6z/sqrt(14))

Substituting the coordinates of the point (1, 2, 3) into the equation, we get:

∇f · (v/|v|) = (4(1)/sqrt(14)) + (-2(2)/sqrt(14)) + (6(3)/sqrt(14))
= 4/sqrt(14) - 4/sqrt(14) + 18/sqrt(14)
= 18/sqrt(14)

Therefore, the directional derivative of f(x, y, z) = x^2 + y^2 + z^2 at the point (1, 2, 3) in the direction of vector v = <2, -1,="" 3=""> is 18/sqrt(14).