The shortest distance (in units) between the lines L_1:(x-1)/2=(y-2)/…

The correct answer is Option (2) → \( \displaystyle\frac{\sqrt{293}}{7} \)

Since \( \frac{3-y}{-3}=\frac{y-3}{3} \), the line \( L_2 \) in standard form is

    \( \displaystyle L_2:\ \frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6} \)

Both lines have direction ratios 2, 3, 6, so they are parallel. For parallel lines \( \vec{r}=\vec{a}_1+\lambda\vec{b} \) and \( \vec{r}=\vec{a}_2+\mu\vec{b} \),

    \( \displaystyle d=\frac{|\vec{b}\times(\vec{a}_2-\vec{a}_1)|}{|\vec{b}|} \)

Here \( \vec{a}_1=\hat{\imath}+2\hat{\jmath}-4\hat{k} \), \( \vec{a}_2=3\hat{\imath}+3\hat{\jmath}-5\hat{k} \), \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+6\hat{k} \).

    \( \displaystyle \vec{a}_2-\vec{a}_1=2\hat{\imath}+\hat{\jmath}-\hat{k} \)

    \( \displaystyle \vec{b}\times(\vec{a}_2-\vec{a}_1)=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 3 & 6\\2 & 1 & -1\end{vmatrix} \)

    \( \displaystyle =\hat{\imath}(-3-6)-\hat{\jmath}(-2-12)+\hat{k}(2-6) \)

    \( \displaystyle =-9\hat{\imath}+14\hat{\jmath}-4\hat{k} \)

    \( \displaystyle |\vec{b}\times(\vec{a}_2-\vec{a}_1)|=\sqrt{81+196+16}=\sqrt{293} \)

    \( \displaystyle |\vec{b}|=\sqrt{4+9+36}=7 \)

    \( \displaystyle d=\frac{\sqrt{293}}{7} \)

Why the other options are incorrect

• Option (1) \( \displaystyle\frac{293}{7} \) — Incorrect. The square root of \( |\vec{b}\times(\vec{a}_2-\vec{a}_1)|^2=293 \) has not been taken.

• Option (3) \( \sqrt{293} \) — Incorrect. \( \sqrt{293} \) is only the numerator; it must be divided by \( |\vec{b}|=7 \).

• Option (4) \( \sqrt{493} \) — Incorrect. The numerator is \( \sqrt{293} \), not \( \sqrt{493} \).

Tip: A line written as \( \frac{3-y}{-3} \) hides the direction ratio 3; rewrite it as \( \frac{y-3}{3} \) before comparing directions. For parallel lines \( \vec{b}_1\times\vec{b}_2=\vec{0} \), so the skew-lines formula cannot be used.