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

\( x=4\sin{2t} \),

\( y=4\cos{2t} \)

and \( z=6t \)

So the displacement of the particle can be written as,

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

\( or,\ \vec{r}= 4\sin{2t}\hat{i}+4\cos{2t}\hat{j}+6t\hat{k} \)

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,

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

\( or,\ \displaystyle{\vec{v}=\frac{d}{dt}[4\sin{2t}\hat{i}+4\cos{2t}\hat{j}+6t\hat{k}]} \)

\( or,\ \vec{v}=4(\cos{2t})\times{2}\hat{i}+4(-\sin{2t})\times{2}\hat{j}+6\hat{k} \)

\( or,\ \vec{v}=8\cos{2t}\hat{i}-8\sin{2t}\hat{j}+6\hat{k} \)

So the magnitude of the velocity is

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

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

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

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

Now the acceleration is given by,

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

\( or,\ \vec{a}=\frac{d}{dt}[8\cos{2t}\hat{i}-8\sin{2t}\hat{j}+6\hat{k}] \)

\( or,\ \vec{a}=8(-\sin{2t})\times{2}\hat{i}-8(\cos{2t})\times{2}\hat{j} \)

\( or,\ \vec{a}=-16\sin{2t}\hat{i}-16\cos{2t}\hat{j} \)

So the magnitude of the acceleration is given by,

\( a=|\vec{a}|=\sqrt{16^2{\sin}^2{2t}+16^2{\cos}^2{2t}} \)

\( or,\ a=\sqrt{16^2({\sin}^2{2t}+{\cos}^2{2t})}=\sqrt{16^2} \)

\( or,\ a=16\ m/{sec}^2 \)