Step-by-step Solution:

Solution: 1. Let the required vector be: $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$. 2. Equal angles with x and y axes: Since the vector makes equal angles with the x and y axes, the direction cosines with respect to these axes must be equal. This means: $$\frac{a}{|\vec{v}|} = \frac{b}{|\vec{v}|}$$ Since $\vec{v}$ is a unit vector, $|\vec{v}| = 1$, so $a = b$. 3. Unit vector condition: Since $\vec{v}$ is a unit vector, we have: $$a^2 + b^2 + c^2 = 1 \quad \text{(i)}$$ 4. Orthogonality condition: The vector $\vec{v}$ is orthogonal to $-\hat{i} + 2\hat{j} + 2\hat{k}$. This means their dot product is zero: $$(a\hat{i} + b\hat{j} + c\hat{k}) \cdot (-\hat{i} + 2\hat{j} + 2\hat{k}) = 0$$ $$-a + 2b + 2c = 0$$ 5. Substitute a = b: Since $a = b$, we can substitute $b$ with $a$ in the orthogonality condition: $$-a + 2a + 2c = 0$$ $$a + 2c = 0$$ $$2c = -a$$ $$c = -\frac{a}{2}$$ 6. Substitute b and c in the unit vector condition (i): Substitute $b = a$ and $c = -\frac{a}{2}$ into $a^2 + b^2 + c^2 = 1$: $$a^2 + a^2 + \left(-\frac{a}{2}\right)^2 = 1$$ $$2a^2 + \frac{a^2}{4} = 1$$ Multiply by 4: $$8a^2 + a^2 = 4$$ $$9a^2 = 4$$ $$a^2 = \frac{4}{9}$$ $$a = \pm \frac{2}{3}$$ 7. Find b and c: Since $b = a$, we have $b = \pm \frac{2}{3}$. Since $c = -\frac{a}{2}$, we have $c = -\frac{1}{2} \left( \pm \frac{2}{3} \right) = \mp \frac{1}{3}$. 8. Write the vector: The vector is: $$\vec{v} = a\hat{i} + b\hat{j} + c\hat{k} = \pm \frac{2}{3} \hat{i} + \pm \frac{2}{3} \hat{j} \mp \frac{1}{3} \hat{k}$$ 9. Factor out $\pm \frac{1}{3}$: $$\vec{v} = \pm \frac{1}{3} (2\hat{i} + 2\hat{j} - \hat{k})$$ Final Answer: The required unit vector is $\pm \frac{1}{3} (2\hat{i} + 2\hat{j} - \hat{k})$.