If the vector $\hat{i}-3 \hat{j}+5 \hat{k}$ bisects the angle between $\hat{a}$ and $-\hat{i}+2 \hat{j}+2 \hat{k}$, where $\hat{a}$ is a unit vector, then:

$\hat{a}=\frac{1}{105}(4 \hat{i}-88 \hat{j}-40 \hat{k})$

We must have

$\lambda(\hat{i}-3 \hat{j}+5 \hat{k})=\hat{a}+\frac{2 \hat{k}+2 \hat{j}-\hat{i}}{3}$

$\Rightarrow 3 \hat{a}=3 \lambda(\hat{i}-3 \hat{j}+5 \hat{k})-(2 \hat{k}+2 \hat{j}-\hat{i})$

$=\hat{i}(3 \lambda+1)-\hat{j}(2+9 \lambda)+\hat{k}(15 \lambda-2)$

$\Rightarrow 3|\hat{a}|=\sqrt{(3 \lambda+1)^2+(2+9 \lambda)^2+(15 \lambda-2)^2}$

$\Rightarrow 9=(3 \lambda+1)^2+(2+9 \lambda)^2+(15 \lambda-2)^2$

$\Rightarrow 315 \lambda^2-18 \lambda=0$

$\Rightarrow \lambda=0, \frac{2}{35}$

For $\lambda=0, \vec{a}=\hat{i}-2 \hat{j}-2 \hat{k}$ (not acceptable)

For $\lambda=\frac{2}{35}, \vec{a}=\frac{41}{105} \hat{i}-\frac{88}{105} \hat{j}-\frac{40}{105} \hat{k}$

Hence (4) is correct answer.