If the shortest distance between the lines...

\left.$$\begin{array}{l}\overline{\mathrm{r}}_1=(\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\alpha(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ \overline{\mathrm{r}}_2=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\beta(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\end{array}$$\right\} \begin{gathered}\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ \overline{\mathrm{a}}_2+\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \mathrm{a}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\end{gathered}$Shortest dist.$=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$\begin{aligned} & \left\lvert\, \frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\right. \\ & |-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{k}+12 \hat{i}+4(2-\lambda) \hat{j}|=13 \\ & |12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13\end{aligned}$$\begin{aligned} & 144+16 \lambda^2+(3 \lambda-6)^2=169 \\ & 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1\end{aligned}$$