Given:
The direction ratios of the line perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) need to be determined.

To find:
The direction cosines of the line.

Solution:
The direction ratios of a line are the coefficients of the variables in the equation of the line. To find the direction cosines, we need to normalize the direction ratios. The direction cosines are the ratios of the direction ratios to the magnitude of the direction ratios.

Step 1: Find the magnitude of the direction ratios.
Let the direction ratios of the line be (a, b, c).
Magnitude = √(a^2 + b^2 + c^2)

For the first line with direction ratios (1, -2, -2):
Magnitude = √(1^2 + (-2)^2 + (-2)^2) = √(1 + 4 + 4) = √9 = 3

For the second line with direction ratios (0, 2, 1):
Magnitude = √(0^2 + 2^2 + 1^2) = √(0 + 4 + 1) = √5

Step 2: Normalize the direction ratios.
Divide each direction ratio by the corresponding magnitude to obtain the direction cosines.

For the first line with direction ratios (1, -2, -2):
Direction cosines = (1/3, -2/3, -2/3)

For the second line with direction ratios (0, 2, 1):
Direction cosines = (0/√5, 2/√5, 1/√5) = (0, 2/√5, 1/√5)

Step 3: Determine the line perpendicular to both lines.
To find the direction cosines of the line perpendicular to both lines, we can take the cross product of their direction cosines.

Cross product = ((-2/3)*(1/√5) - (-2/3)*(2/√5), (-2/3)*(1/√5) - (1/3)*(0), (1/3)*(2/√5) - (0)*(2/√5))

Simplifying the cross product:
= (-2/3)*(1/√5) - (-4/3)*(1/√5), (-2/3)*(1/√5) - 0, (2/3)*(2/√5) - 0)
= (-6/3√5 + 4/3√5, -2/3√5, 2/3√5)
= (-2/3√5, -2/3√5, 2/3√5)

Therefore, the direction cosines of the line perpendicular to the given lines are (-2/3√5, -2/3√5, 2/3√5).

Conclusion:
The correct option is D) 2/3, -1/3, 2/3.