$\frac{x-0}{1}=\frac{y+2 \lambda}{1}=\frac{z-\lambda}{1}$
Shortest distance $=\frac{\left( a _{2}- a _{1}\right) \cdot\left( b _{1} \times b _{2}\right)}{\left| b _{1} \times b _{2}\right|}$
$b _{1} \times b _{2}=\left|\begin{array}{ccc} i & j & k \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 1\end{array}\right|$
$=\hat{ i }\left(\frac{1}{2}+\frac{1}{2}\right)-\hat{ j }\left(1+\frac{1}{2}\right)+\hat{ k }\left(1-\frac{1}{2}\right)$
$=\hat{ i }-\frac{3}{2} \hat{ j }+\frac{\hat{ k }}{2}=\frac{2 \hat{ i }-3 \hat{ j }+\hat{ k }}{2}$
$\frac{b_{1} \times b_{2}}{\left|b_{1} \times b_{2}\right|}=\frac{2 \hat{i}-3 \hat{j}+\hat{k}}{\sqrt{14}}$
$\frac{\left(a_{2}-a_{1}\right) \cdot\left(b_{1} \times b_{2}\right)}{\left|b_{1} \times b_{2}\right|}=\left(-\lambda \hat{i}+\left(-2 \lambda+\frac{1}{2}\right)+\lambda \hat{k}\right)$
$\quad\quad\quad\quad\quad\quad\quad\left(\frac{2 \hat{i}-3 \hat{j}+\hat{k}}{\sqrt{14}}\right)$
$=\left|\frac{-2 \lambda+6 \lambda-\frac{3}{2}+\lambda}{\sqrt{14}}\right|=\frac{\sqrt{7}}{2 \sqrt{2}}$
$\left|5 \lambda-\frac{3}{2}\right|=\frac{7}{2}$
$5 \lambda=\frac{3}{2} \pm \frac{7}{2}$
$5 \lambda=5,-2$
$\lambda=1,-\frac{2}{5}$
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