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(C) Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ be a unit vector,so $a_1^2 + a_2^2 + a_3^2 = 1$. Let $\vec{b} = 2 \hat{i} - \hat{j} + 2 \h

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$\frac{1}{\sqrt{155}}(7 \hat{i}+9 \hat{j}+5 \hat{k})$

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(C) Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ be a unit vector,so $a_1^2 + a_2^2 + a_3^2 = 1$. Let $\vec{b} = 2 \hat{i} - \hat{j} + 2 \hat{k}$,$\vec{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$,and $\vec{d} = \hat{k}$. The projections are equal,so $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{\vec{a} \cdot \vec{d}}{|\vec{d}|}$. Calculating magnitudes: $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = 3$,$|\vec{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = 3$,and $|\vec{d}| = 1$. Thus,$\frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = a_3$. From $\frac{2a_1 - a_2 + 2a_3}{3} = a_3$,we get $2a_1 - a_2 - a_3 = 0$. From $\frac{a_1 + 2a_2 - 2a_3}{3} = a_3$,we get $a_1 + 2a_2 - 5a_3 = 0$. Solving these equations,we find $a_1 = \frac{7}{5}a_3$ and $a_2 = \frac{9}{5}a_3$. Substituting into $a_1^2 + a_2^2 + a_3^2 = 1$: $(\frac{7}{5}a_3)^2 + (\frac{9}{5}a_3)^2 + a_3^2 = 1 \implies \frac{49+81+25}{25}a_3^2 = 1 \implies a_3^2 = \frac{25}{155} \implies a_3 = \frac{5}{\sqrt{155}}$. Thus,$\vec{a} = \frac{1}{\sqrt{155}}(7 \hat{i} + 9 \hat{j} + 5 \hat{k})$.

If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$,$\hat{i}+2 \hat{j}-2 \hat{k}$,and $\hat{k}$ are equal,then a unit vector along $\vec{a}$ is:

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