Answer
The angle between the direction of the two vectors is $~~127^{\circ}$
Work Step by Step
We can express the object's velocity in unit-vector notation:
$v = (-6.00~m/s)~\hat{i}+ (3.00~m/s)~\hat{j}+ (3.00~m)~\hat{k}$
We can express the force in unit-vector notation:
$F = (6.00~N)~\hat{i}-(8.00~N)~\hat{j}+(4.00~N)~\hat{k}$
We can find the magnitude of $v$:
$\vert v \vert = \sqrt{(-6.00~m/s)^2+(3.00~m/s)^2+(3.00~m/s)^2}$
$\vert v \vert = \sqrt{54.0}~m/s$
We can find the magnitude of $F$:
$\vert F \vert = \sqrt{(6.00~N)^2+(-8.00~N)^2+(4.00~N)^2}$
$\vert F \vert = \sqrt{116}~N$
We can find the angle between the two vectors:
$cos~\theta = \frac{v\cdot F}{\vert v \vert \cdot \vert F \vert}$
$cos~\theta = \frac{(-6.00)(6.00)+(3.00)(-8.00)+(3.00)(4.00)}{(\sqrt{54.0})~(\sqrt{116})}$
$cos~\theta = \frac{-36.0-24.0+12.0}{(\sqrt{54.0})~(\sqrt{116})}$
$cos~\theta = \frac{-48.0}{(\sqrt{54.0})~(\sqrt{116})}$
$cos~\theta = -0.6065$
$\theta = cos^{-1}~(-0.6065)$
$\theta = 127^{\circ}$
The angle between the direction of the two vectors is $~~127^{\circ}$
