Answer
$4.19\;\rm m/s$, $17.35^\circ$ north of east.
Work Step by Step
We assume that the momentum is conserved just before and just after the collision.
$$p_i=p_f$$
$$m_1v_{i1}+m_2v_{i2}+m_3v_{i3}=(m_1+m_2 +m_3)v_f$$
Since the three cars entangled and slide as one body.
Noting that $m_1=2100$ kg which is our unlucky truck, $m_2=1200$ kg which is the one that is traveling north, and $m_3=1500$ kg which is the one that hits our track from the rear.
Thus,
$$2100v_{i1}+1200v_{i2}+1500v_{i3}=(2100+1200 +1500)v_f$$
$$2100v_{i1}+1200v_{i2}+1500v_{i3}=4800\;v_f$$
$$21 v_{i1}+12 v_{i2}+15 v_{i3}=48 \;v_f\tag 1$$
Now we need to write the initial velocities of the three vehicles in components form in which the positive $x$-direction is toward the east and the positive $y$-direction is toward the north.
$$v_{i1}=(2\;\hat i+0\;\hat j)\rm m/s$$
$$v_{i2}=(0\;\hat i+5\;\hat j)\rm m/s$$
$$v_{i3}=(10\;\hat i+0\;\hat j)\rm m/s$$
Plugging these last 3 formulas into (1) and solving for $v_f$;
$$21 (2\;\hat i+0\;\hat j) +12 (0\;\hat i+5\;\hat j)+15 (10\;\hat i+0\;\hat j)=48 \;v_f$$
$$ 42\;\hat i +60\;\hat j +150\;\hat i =48 \;v_f$$
Thus,
$$v_f=(4\;\hat i+1.25\;\hat j)\;\rm m/s$$
The speed of the resulting ball is given by
$$|v_f|=\sqrt{v_x^2+v_y^2}=\sqrt{4^2+1.25^2} =\color{red}{\bf 4.19}\;\rm m/s $$
And its direction is in the first quadrant since it has a positive $x$-component and a positive $y$-component as well.
And it is given by
$$\theta=\tan^{-1}\left( \dfrac{v_y}{v_x} \right)=\tan^{-1}\left( \dfrac{1.25}{4} \right)$$
$$\theta=\color{red}{\bf 17.35^\circ}\tag{north of east}$$
