91Ó°ÊÓ

We are given,

$\mathbf{r}\left(t\right)=⟨t,5\sin 3t,5\cos 3tâŸ\copyright ,t=\frac{\mathrm{Ï€}}{6}$

Finding the binormal vector,

$$\begin{gathered} r\left(t\right)=⟨t,5\sin 3t,5\cos 3tâŸ\copyright \\ {r}^{\text{'}}\left(t\right)=⟨1,15\cos 3t,-15\sin 3tâŸ\copyright \\ ||r^{\text{'}}\left(t\right)||=\sqrt{1+(15\cos 3t{)}^2+(-15\sin 3t{)}^2} \\ =\sqrt{1+225\left({\cos }^{2}3t+{\sin }^{2}3t\right)} \\ =\sqrt{226} \\ T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨1,15\cos 3t,-15\sin 3tâŸ\copyright }{\sqrt{226}} \end{gathered}$$

At $t=\frac{\mathrm{Ï€}}{6}$,

role="math" localid="1649782675512"

So,

At $t=\frac{\mathrm{Ï€}}{6}$,

Thus, the principal unit vector is $⟨0,-1,0âŸ\copyright$.

Now,

The binormal vector is given by,

The equation of the osculating at $r\left(t=\frac{\mathrm{Ï€}}{6}\right)$is defined by,

Hence, the equation is $30x+2z=5\mathrm{Ï€}$.