A sailor went to sea, sea, sea to see what he could see, see, see

Let the Velocity of sailor be $V_{B}$

So $V_{ B }=8\hat { i }$

$V_{ SB }=12\hat { j } -2\hat { k }$
( $V_{SB}$ is velocity of Submarine wrt sailor)

$V_{ HS }=5\hat { k } -4\hat { i }$ ( $V_{HS}$ is velocity of Helicopter wrt Submarine)

Now $V_{ SB } =V_{S}-V_{B}$

and $V_{ HS } =V_{H} -V_{S}$

Adding these equations gives

$V_{ SB }+V_{ HS }=V_{H}-V_{B}$ and this equal to $V_{HB}$ i.e velocity helicopter with respect to the boat.

$V_{ SB }+V_{ HS }=-4\hat { i } +12\hat { j } +3\hat { k }$

So its speed is $\sqrt { 144+16+9 }$ i.e. 13m/s.

Define a suitable three dimensional coordinate system $xyz$ , where positive $x$ is North, positive $y$ is West, and positive $z$ is upwards (higher altitude).

Now, represent the speed components of each vehicle in vector form: Boat: $\overrightarrow{B} =[0,-8,0]$ ; Submarine: $\overrightarrow{S} =[12,0,-2]$ ; Helicopter: $\overrightarrow{H} =[0,4,5]$ .

The vector speed of the helicopter with respect to the boat is the vector $\overrightarrow {BH} = [0,12,5]$ . The magnitude of this vector is $\sqrt{0^2+12^2+5^2}= \boxed{13}$ .