If the vector $-\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}$ bisects the angles between the vector $\overline{\mathrm{c}}$ and the vector $3\overline{\mathrm{i}}+4\overline{\mathrm{j}}$, then the unit vector in the direction of $\overline{\mathrm{c}}$ is

$\mathrm{Let}\overrightarrow{\mathrm{c}}=\mathrm{x}\hat{i}+y\hat{j}+z\hat{k}$$\text{ where }{x}^2+y^2+z^2=1$

$\text{ And unit vector along }3\hat{i}+4\hat{j}=\frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^2}}=\frac{3\hat{i}+4\hat{j}}{5}$

$The angle biseector of \hat{a} , \hat{b} is\overrightarrow{ r}=t\left(\hat{a} + \hat{b}\right)$

$$\begin{gathered} r=t\left(x \hat{i}+y \hat{j}+z\hat{k}+\frac{3\hat{i}+4\hat{j}}{5}\right) \\ \mathrm{r}=\frac{\mathrm{t}}{5}\left[\right(5\mathrm{x}+3)\hat{i}+(5\mathrm{y}+4)\widehat{ j}+5z\hat{k}) \\ \text{ But the bisector is given by }-\hat{i}+\hat{j}-\hat{k} \end{gathered}$$

$\begin{array}{l}\frac{t}{5}(5x+3)=-1\Rightarrow x=-\frac{5+3t}{5t},\\ \frac{t}{5}(5y+4)=1\Rightarrow y=\frac{5-4t}{5t},\\ \frac{t}{5}\left(5z\right)=-1\Rightarrow z=-\frac{1}{t}\end{array}$

$\text{ Put all the values in equation (1), we get }$

$\begin{array}{r}{\left(-\frac{5+3t}{5t}\right)}^2+{\left(\frac{5-4t}{5t}\right)}^2+{\left(-\frac{1}{t}\right)}^2=1\\ \frac{25+9t^2+30t+25+16t^2-40t+25}{25t^2}=1\\ 25t^2-10t+75=25t^2\\ t=7.5\end{array}$

$$\begin{gathered} \begin{array}{r}x=-\frac{5+3\times 7.5}{5\times 7.5}=-\frac{11}{15},y=\frac{5-4\times 7.5}{5\times 7.5}=-\frac{10}{15},z=-\frac{1}{7.5}=-\frac{2}{15}\end{array} \\ \overrightarrow{\mathrm{c}}=\frac{1}{15}(-11\hat{i}-10\hat{j}-2\hat{k}) \end{gathered}$$