The motion of the particle is described by the three equations, 

\( x=2\sin{3t} \),

\( y=2\cos{3t} \)

and \( z=8t \)

So the displacement of the particle can be written as,

\( \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \)

\( or,\ \vec{r}= 2\sin{3t}\hat{i}+2\cos{3t}\hat{j}+8t\hat{k} \)

where, \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \) are the unit vectors in three dimensional Cartesian co-ordinate system along the X-axis, Y-axis and Z-axis respectively.

Again the velocity is the time derivative of the displacement vector, so the velocity is given by,

\( \displaystyle{\vec{v}=\frac{d\vec{r}}{dt}} \)

\( or,\ \displaystyle{\vec{v}=\frac{d}{dt}[2\sin{3t}\hat{i}+2\cos{3t}\hat{j}+8t\hat{k}]} \)

\( or,\ \vec{v}=2(\cos{3t})\times{3}\hat{i}+2(-\sin{3t})\times{3}\hat{j}+8\hat{k} \)

\( or,\ \vec{v}=6\cos{3t}\hat{i}-6\sin{3t}\hat{j}+8\hat{k} \)

So the magnitude of the velocity is

\( v=|\vec{v}|=\sqrt{6^2{\cos}^2{3t}+6^2{\sin}^2{3t}+8^2} \)

\( or,\ v=\sqrt{6^2({\cos}^2{3t}+{\sin}^2{3t})+8^2} \)

\( or,\ v=\sqrt{6^2+8^2} \)

\( or,\ v=\sqrt{100}=10\ m/sec \)

Now the acceleration is given by,

\( \vec{a}=\frac{d\vec{v}}{dt} \)

\( or,\ \vec{a}=\frac{d}{dt}[6\cos{3t}\hat{i}-6\sin{3t}\hat{j}+8\hat{k}] \)

\( or,\ \vec{a}=6(-\sin{3t})\times{3}\hat{i}-6(\cos{3t})\times{3}\hat{j} \)

\( or,\ \vec{a}=-18\sin{3t}\hat{i}-18\cos{3t}\hat{j} \)

So the magnitude of the acceleration is given by,

\( a=|\vec{a}|=\sqrt{18^2{\sin}^2{3t}+18^2{\cos}^2{3t}} \)

\( or,\ a=\sqrt{18^2({\sin}^2{3t}+{\cos}^2{3t})}=\sqrt{18^2} \)

\( or,\ a=18\ m/{sec}^2 \)