Sketch the vector diagram.

To illustrate the given information, we draw a coordinate system with North pointing upwards, South downwards, East to the right, and West to the left. All vectors originate from the central point (origin).

• Vector 1 ($V_1 = 3 \text{ ms}^{-1}$ N45W): Draw an arrow from the origin into the second quadrant, exactly halfway between the North (positive y-axis) and West (negative x-axis) directions. This vector makes an angle of $45^\circ$ with the North axis towards the West, or $135^\circ$ counter-clockwise from the positive East (x-axis).
• Vector 2 ($V_2 = 12 \text{ ms}^{-1}$ W): Draw an arrow from the origin directly along the negative x-axis (West direction).
• Vector 3 ($V_3 = 5 \text{ ms}^{-1}$ S): Draw an arrow from the origin directly along the negative y-axis (South direction).

The resultant vector would be drawn from the origin to the final position if these vectors were added head-to-tail.

Let the positive x-axis be East and the positive y-axis be North.

For $V_1 = 3 \text{ ms}^{-1}$ N45W:
The angle from the positive x-axis is $90^\circ + 45^\circ = 135^\circ$.

$$V_{1x} = 3 \cos(135^\circ) = 3 \left(-\frac{\sqrt{2}}{2}\right) = -1.5\sqrt{2} \text{ ms}^{-1}$$ $$V_{1y} = 3 \sin(135^\circ) = 3 \left(\frac{\sqrt{2}}{2}\right) = 1.5\sqrt{2} \text{ ms}^{-1}$$

For $V_2 = 12 \text{ ms}^{-1}$ W:
The angle from the positive x-axis is $180^\circ$.

$$V_{2x} = 12 \cos(180^\circ) = 12(-1) = -12 \text{ ms}^{-1}$$ $$V_{2y} = 12 \sin(180^\circ) = 12(0) = 0 \text{ ms}^{-1}$$

For $V_3 = 5 \text{ ms}^{-1}$ S:
The angle from the positive x-axis is $270^\circ$.

$$V_{3x} = 5 \cos(270^\circ) = 5(0) = 0 \text{ ms}^{-1}$$ $$V_{3y} = 5 \sin(270^\circ) = 5(-1) = -5 \text{ ms}^{-1}$$

$$R_x = V_{1x} + V_{2x} + V_{3x} = -1.5\sqrt{2} - 12 + 0 = -12 - 1.5\sqrt{2} \text{ ms}^{-1}$$ $$R_y = V_{1y} + V_{2y} + V_{3y} = 1.5\sqrt{2} + 0 - 5 = -5 + 1.5\sqrt{2} \text{ ms}^{-1}$$

Using $\sqrt{2} \approx 1.414$:

$$R_x \approx -12 - 1.5(1.414) = -12 - 2.121 = -14.121 \text{ ms}^{-1}$$ $$R_y \approx -5 + 1.5(1.414) = -5 + 2.121 = -2.879 \text{ ms}^{-1}$$

The magnitude $R$ is given by:

$$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(-14.121)^2 + (-2.879)^2}$$ $$R = \sqrt{199.39 + 8.288}$$ $$R = \sqrt{207.678}$$ $$R \approx 14.41 \text{ ms}^{-1}$$

Since both $R_x$ and $R_y$ are negative, the resultant vector lies in the third quadrant (South-West).
Let $\phi$ be the angle the resultant makes with the negative x-axis (West).

$$\tan \phi = \left|\frac{R_y}{R_x}\right| = \left|\frac{-2.879}{-14.121}\right| = \frac{2.879}{14.121} \approx 0.20387$$ $$\phi = \arctan(0.20387) \approx 11.53^\circ$$

The direction can be stated as $11.53^\circ$ South of West (W11.53S).

The resultant of the three vectors is $\boxed{\text{14.41 ms}^{-1} \text{ at } 11.53^\circ \text{ South of West}}$.