$\begin{array}{ll} & x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}) \\ \therefore \quad & \frac{\mathrm{d} x}{\mathrm{dt}}=2(-\sin \mathrm{t}+\sin \mathrm{t}+\mathrm{t} \cos \mathrm{t})=2 \mathrm{t} \cos \mathrm{t} \\ & y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}) \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=2(\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t})=2 \mathrm{t} \sin \mathrm{t} \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{2 \mathrm{t} \sin \mathrm{t}}{2 \mathrm{t} \cos \mathrm{t}}=\mathrm{tan} \mathrm{t} \\ & \text { Slope of normal }=-\frac{1}{\frac{\mathrm{d} y}{\mathrm{~d} x}}=-\frac{1}{\tan \mathrm{t}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}}\end{array}$
$\therefore \quad$ Equation of the normal is
$$
\begin{aligned}
& y-2(\sin t-t \cos t)=-\frac{\cos t}{\sin t}[x-2(\cos t+t \sin t)] \\
& \Rightarrow y \sin \mathrm{t}-2 \sin ^2 \mathrm{t}+2 \mathrm{t} \sin \mathrm{t} \cos \mathrm{t} \\
& =-x \cos t+2 \cos ^2 t+2 t \sin t \cos t \\
& \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2\left(\sin ^2 \mathrm{t}+\cos ^2 \mathrm{t}\right) \\
& \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2 \\
& \therefore \quad \text { Distance from origin }=\left|\frac{-2}{\sqrt{\cos ^2 \mathrm{t}+\sin ^2 \mathrm{t}}}\right|=2 \text { units } \\
&
\end{aligned}
$$