$$ \begin{aligned} & \text { SHORTEST distance } \frac{\left|\left(a_2-a_1\right) \cdot\left(b_1 \times b_2\right)\right|}{\left|b_1 \times b_2\right|} \\ & a_1=(1,2,3) \\ & a_2=(2,4,5) \\ & \bar{b}_2=2 \hat{i}+3 \hat{j}+\lambda \hat{k} \\ & \bar{b}_2=\hat{i}+4 \hat{j}+5 \hat{k} \\ & \text { S.D. }=\frac{\mid(2-1) \hat{i}+(4-2) \hat{j}+(5-3) \hat{k}) \cdot\left(\bar{b}_1 \times \bar{b}_2\right) \mid}{\left|b_1 \times b_2\right|} \\ & \bar{b}_1 \times \bar{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5 \end{array}\right| \\ & =\hat{i}(15-4 \lambda)+\hat{j}(\lambda-10)+\hat{k}(5) \\ & =(15-4 \lambda) \hat{i}+(\lambda-10) \hat{j}+5 \hat{k} \\ & =\left|\bar{b}_1 \times \bar{b}_2\right|-\sqrt{(15-4 \lambda)^2+(\lambda-10)^2+25} \end{aligned} $$ Now $$ \begin{aligned} & \text { S.D. }=\frac{\mid \hat{i}+2 \hat{j}+2 \hat{k}) \cdot[(15-4 \lambda) \hat{i}+(\lambda-10) \hat{j}+5 \hat{k}] \mid}{\sqrt{(15-4 \lambda)^2+(\lambda-10)^2+25}} \\ & \frac{|15-4 \lambda+2 \lambda-20+10|}{\sqrt{(15-4 \lambda)^2+(\lambda-10)^2+25}}=\frac{1}{\sqrt{3}} \end{aligned} $$ square both side $$ \begin{aligned} & 3(5-2 \lambda)^2=225+16 \lambda^2-120 \lambda+\lambda^2+100-20 \lambda+25 \\ & 12 \lambda^2+75-60 \lambda=17 \lambda^2-140 \lambda+350 \\ & 5 \lambda^2-80 \lambda+275=0 \\ & \lambda^2-16 \lambda+55=0 \\ & (\lambda-5)(\lambda-11)=0 \\ & \Rightarrow \lambda=5,11 \end{aligned} $$