Question:

Step 1: Let the required vector be $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$. Since $\mathbf{r}$ is a unit vector, its magnitude is 1: $$x^2 + y^2 + z^2 = 1 \quad (*)$$ Step 2: The vector $\mathbf{r}$ is perpendicular to $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$. The dot product of two perpendicular vectors is zero: $$\mathbf{r} \cdot (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 0$$ $$x(1) + y(2) + z(2) = 0$$ $$x + 2y + 2z = 0 \quad (**)$$ Step 3: The vector $\mathbf{r}$ makes equal angles with the X and Y-axes. For a unit vector $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, the direction cosines are $x$, $y$, and $z$. If the angles with the X and Y-axes are equal, then their cosines are equal in magnitude: $$|\cos\alpha_x| = |\cos\alpha_y| \Rightarrow |x| = |y|$$ This implies two possibilities: $x=y$ or $x=-y$. Step 4: Case 1: $x=y$. Substitute $y=x$ into equation $(**)$: $$x + 2x + 2z = 0$$ $$3x + 2z = 0 \Rightarrow z = -\frac{3}{2}x$$ Substitute $y=x$ and $z=-\frac{3}{2}x$ into equation $(*)$: $$x^2 + x^2 + \left(-\frac{3}{2}x\right)^2 = 1$$ $$2x^2 + \frac{9}{4}x^2 = 1$$ $$\frac{8x^2 + 9x^2}{4} = 1$$ $$\frac{17x^2}{4} = 1 \Rightarrow x^2 = \frac{4}{17} \Rightarrow x = \pm \frac{2}{\sqrt{17}}$$ If $x = \frac{2}{\sqrt{17}}$, then $y = \frac{2}{\sqrt{17}}$ and $z = -\frac{3}{2}\left(\frac{2}{\sqrt{17}}\right) = -\frac{3}{\sqrt{17}}$. Thus, $\mathbf{r}_1 = \frac{1}{\sqrt{17}}(2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k})$. If $x = -\frac{2}{\sqrt{17}}$, then $y = -\frac{2}{\sqrt{17}}$ and $z = -\frac{3}{2}\left(-\frac{2}{\sqrt{17}}\right) = \frac{3}{\sqrt{17}}$. Thus, $\mathbf{r}_2 = \frac{1}{\sqrt{17}}(-2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k})$. Step 5: Case 2: $x=-y$. Substitute $y=-x$ into equation $(**)$: $$x + 2(-x) + 2z = 0$$ $$-x + 2z = 0 \Rightarrow z = \frac{1}{2}x$$ Substitute $y=-x$ and $z=\frac{1}{2}x$ into equation $(*)$: $$x^2 + (-x)^2 + \left(\frac{1}{2}x\right)^2 = 1$$ $$2x^2 + \frac{1}{4}x^2 = 1$$ $$\frac{8x^2 + x^2}{4} = 1$$ $$\frac{9x^2}{4} = 1 \Rightarrow x^2 = \frac{4}{9} \Rightarrow x = \pm \frac{2}{3}$$ If $x = \frac{2}{3}$, then $y = -\frac{2}{3}$ and $z = \frac{1}{2}\left(\frac{2}{3}\right) = \frac{1}{3}$. Thus, $\mathbf{r}_3 = \frac{1}{3}(2\mathbf{i} - 2\mathbf{j} + \mathbf{k})$. If $x = -\frac{2}{3}$, then $y = \frac{2}{3}$ and $z = \frac{1}{2}\left(-\frac{2}{3}\right) = -\frac{1}{3}$. Thus, $\mathbf{r}_4 = \frac{1}{3}(-2\mathbf{i} + 2\mathbf{j} - \mathbf{k})$. Step 6: The possible unit vectors satisfying the given conditions are: $$\frac{1}{\sqrt{17}}(2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}), \quad \frac{1}{\sqrt{17}}(-2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}),$$ $$\frac{1}{3}(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}), \quad \text{and} \quad \frac{1}{3}(-2\mathbf{i} + 2\mathbf{j} - \mathbf{k}).$$