Answer
An equation of the osculating plane of the curve:
$$
x=\sin 2t, \ \ y=t, \ \ z=\cos 2t
$$
at the point $(0,\pi,1)$ is:
$$
x-2 y+2 \pi=0
$$
Work Step by Step
We have the parametric equations of the curve:
$$
x=\sin 2t, \ \ y=t, \ \ z=\cos 2t
$$
So, the curve is:
$$
\mathrm{r}(t)=\left\langle \sin 2t, t, \cos 2t \right\rangle
$$
and the point $(0,\pi, 1)$ corresponds to $ t=\pi. $
Now, we first compute the ingredients needed :
$$
r^{\prime}(t) =\left\langle 2 \cos 2t, 1, -2\sin2t \right\rangle
$$
$\Rightarrow$
$$
\begin{aligned}
\left|r^{\prime}(t)\right|&=\sqrt{(2 \cos 2t)^{2}+(1)^{2}+(-2\sin2t )^{2}}\\
&=\sqrt{1+4(\cos2t)^{2}+(\sin2t )^{2}}\\
&=\sqrt {5}
\end{aligned}
$$
Then, the unit tangent vector $T(t)$ is given by:
$$
\begin{aligned}
T(t) &=\frac{r^{\prime}(t)}{\left|r^{\prime}(t)\right|}\\
&=\frac{1}{\sqrt {5} }\left\langle 2 \cos 2t, 1, -2\sin2t \right\rangle\\
& \ \ \ \ \ \text{corresponds to} \ \ t=\pi ,\\\
&=\frac{1}{\sqrt {5} }\left\langle 2 \cos 2(\pi), 1, -2\sin2(\pi) \right\rangle\\
&=\frac{1}{\sqrt {5} }\left\langle 2, 1, 0 \right\rangle\\
\end{aligned}
$$
Then, we can find:
$$
\begin{aligned}
\mathbf{T}^{\prime}(t)&=\frac{1}{\sqrt {5} }\left\langle -4 \sin 2t, 0, -4\cos 2t \right\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
\left|\mathrm{T}^{\prime}(t)\right|& =\frac{1}{\sqrt {5} }\sqrt{\left(-4 \sin 2t \right)^{2}+(0)^{2}+(-4\cos 2t)^{2}}\\
&=\frac{4}{\sqrt {5} }
\end{aligned}
$$
Then, the principal unit normal vector $N(t)$ is given by:
$$
\begin{aligned}
\mathrm{N}(t)&=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|} \\
&=\frac{\frac{1}{\sqrt {5} }\left\langle -4 \sin 2t, 0, -4\cos 2t \right\rangle}{\frac{4}{\sqrt {5} }} \\
&=\left\langle - \sin 2t, 0, -\cos 2t \right\rangle\\
& \ \ \ \ \ \text{corresponds to} \ \ t=\pi ,\\\
\mathrm{N}(\pi) &=\left\langle - \sin 2(\pi), 0, -\cos 2(\pi) \right\rangle\\
&=\left\langle 0, 0, -1 \right\rangle
\end{aligned}
$$
The binormal vector $\mathbf{B}(\pi)$ is given by:
$$
\begin{aligned}
\mathbf{B}(\pi)=\mathbf{T}(\pi) \times \mathbf{N}(\pi)&=\frac{1}{\sqrt {5} }\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 0 \\
0, & 0 & -1 \\
\end{array}\right]\\
&=\frac{1}{\sqrt {5} } \left\langle -1,2 ,0 \right\rangle.\\
\end{aligned}
$$
A simpler normal vector is $ \left\langle -1,2 ,0 \right\rangle $, so an equation of the osculating plane is
$$
-1(x-0)+2(y-\pi)+0(z-1)=0
$$
or
$$
x-2 y+2 \pi=0
$$
