If the vector $-\hat i +\hat j -\hat k$ bisects the angle between the vector c ́and the vector $3\hat i +4\hat j$, then the unit vector in the direction of $\vec c$ is

$-\frac{1}{15}(11\hat i+10\hat j+2\hat k)$

Let $x\hat i+y\hat j+z\hat k$ be the unit vector along $\vec c$.

Since $-\hat i+\hat j-\hat k$ bisects the angle between $\vec c$ and $3\hat i+\hat j$.

$∴λ(-\hat i+\hat j-\hat k) = (x\hat i+y\hat j+z\hat k) +\frac{3\hat i+4\hat j}{5}$

$⇒x+\frac{3}{5}=-λ,y+\frac{4}{5}=λ$ and $z=-λ$

Now, $x^2+ y^2+z^2=1$   [$∵x\hat i+y\hat j+z\hat k$ is a unit vector]

$⇒(-λ+-\frac{3}{5})^2+(λ-\frac{4}{5})^2+λ^2=1⇒ λ=0$ or, $λ=\frac{2}{15}$

But, $λ≠0$. Because $λ = 0$ implies that the given vectors are parallel.

$∴λ =\frac{2}{15}⇒x=-\frac{11}{15},y=-\frac{10}{15}$ and $z =-\frac{2}{15}$

Hence, $x\hat i+y\hat j+z\hat k=-\frac{1}{15}(11\hat i+10\hat j+2\hat k)$.