Let vector $\mathbf{a}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$
Now, vector bisector of angle between $\mathbf{a}$ and $6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}$ is
$\begin{aligned} & \lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|}+\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{\sqrt{6^2+8^2}}\right)=\lambda\left(\frac{\mathbf{a}}{|\mathbf{a}|}+\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}}{10}\right) \\ & =19 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\end{aligned}$
(given)
$\Rightarrow \frac{\mathbf{a}}{|\mathbf{a}|}=\frac{19 \hat{\mathbf{i}}+22 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}}{\lambda}-\left(\frac{3}{5} \hat{\mathbf{i}}+\frac{4}{5} \hat{\mathbf{j}}\right)$
$=\left(\frac{19}{\lambda}-\frac{3}{5}\right) \hat{\mathbf{i}}+\left(\frac{22}{\lambda}-\frac{4}{5}\right) \hat{\mathbf{j}}+\frac{5}{\lambda} \hat{\mathbf{k}}$
$\because \frac{\mathbf{a}}{|\mathbf{a}|}$ is unit vector along $\mathbf{a}$, so
$\begin{aligned} & \left(\frac{19}{\lambda}-\frac{3}{5}\right)^2+\left(\frac{22}{\lambda}-\frac{4}{5}\right)^2+\left(\frac{5}{\lambda}\right)^2=1 \\ & \frac{361}{\lambda^2}+\frac{484}{\lambda^2}+\frac{25}{\lambda^2}+\frac{9}{25}+\frac{16}{25}-\frac{114}{5 \lambda}-\frac{176}{5 \lambda}=1 \\ & \Rightarrow \frac{870}{\lambda^2}-\frac{290}{5 \lambda}=0 \Rightarrow \lambda=15 \\ & \text { So, } \frac{\mathbf{a}}{|\mathbf{a}|}=\left(\frac{19}{15}-\frac{3}{5}\right) \hat{\mathbf{i}}+\left(\frac{22}{15}-\frac{4}{5}\right) \hat{\mathbf{j}}+\frac{\hat{\mathbf{k}}}{3} \\ & =\frac{10}{15} \hat{\mathbf{i}}+\frac{10}{15} \hat{\mathbf{j}}+\frac{\hat{\mathbf{k}}}{3}=\frac{1}{3}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}$
Hence, option (b) is correct.