Forces of magnitudes 5 and 3 units acting in the directions $6\hat i + 2\hat j + 3\hat k$ and $3\hat i -2\hat j+6\hat k$ respectively act on a particle which is displaced from the point (2, 2, -1) to (4, 3, 1). The work done by the forces, is

$\frac{148}{7}$ units

Let $\vec F$ be the resultant force and $\vec d$ be the displacement vector. Then,

$\vec F=5\frac{(6\hat i+2\hat j+3\hat k)}{\sqrt{36+4+9}}+3\frac{(3\hat i-2\hat j+6\hat k)}{\sqrt{9+4+36}}=\frac{1}{7}(39\hat i+4\hat j+33\hat k)$

and $\vec d=(4\hat i + 3\hat j+\hat k)-(2\hat i +2\hat j-\hat k) = 2\hat i +\hat j+2\hat k$

∴ Total work done = $\vec F. \vec d$

⇒ Total work done = $\frac{1}{7}(39\hat i+4\hat j+33\hat k).(2\hat i +\hat j+2\hat k)$

⇒ Total work done =$\frac{1}{7}(78+4+66)=\frac{148}{7}$ units