The distance of the point \( (1,2,1) \) from the line \( \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2} \) is

\(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda\)
\(x-2 d+1, y=\lambda+2, z=2 d+3\)
\(\begin{aligned} \overrightarrow{P A}=& {[(2 \lambda+1) \hat{\jmath}+(\lambda+2) \hat{\jmath}+(2 \hat{\lambda}+3) \hat{k})] } \\ &-[\hat{\jmath}+2 \hat{\jmath}+\hat{k}] \end{aligned}\)
\(\left[2 \lambda \hat{\imath} \vdash \lambda \hat{\lambda}+(2 \lambda+2) \sum \vec{k}\right]\)
\(\overrightarrow{P A} \perp L\)
\(\therefore \overrightarrow{P A} \cdot L=0\)
\([(2 \lambda \hat{\jmath}+\lambda \hat{\jmath})+(2 \lambda+2) \vec{p}] \cdot[2 \hat{\imath}+\hat{\jmath}+2 \hat{k}]=0\)
\([(2 \lambda \times 2) \hat{i}+(\lambda \times 1) \hat{\jmath}+(2 \lambda+2) \times 2 \hat{k}]=0\)
\([4 \lambda+\lambda+4 d+u]=0\)
\(d=-4 / 9\)
\(A=[2 \lambda+1, \lambda+2,2 \lambda+3]\)
\(A=\left[\frac{1}{3}, \frac{14}{3}, \frac{19}{9}\right]\)
\(D=\sqrt{(1-1 / 9)^{2}+\left(\frac{2+14}{9}\right)^{2}+\left(1-\frac{19}{9}\right)^{2}}\)
\(D=\sqrt{\frac{(4}{81}+\frac{16}{81}+\frac{100}{81}}=\frac{2 \sqrt{5}}{3}\)