Q.The lines $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ and $\dfrac{x-1}{-2}=\dfrac{y-2}{-4}=\dfrac{z-3}{6}$ are
(A) perpendicular
(B) inrersecting
(C) skew
(D) coincident

[!TLDR] Check parallel first (DR ratios), then check coplanarity/common point.

DRs $(1,2,3)$ and $(-2,-4,6)$ are not proportional (not parallel). Point $(1,2,3)$ satisfies both line 1 (at parameter $1$) and is the given point of lin …

Line 1: $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$, through the origin with direction $(1,2,3)$.

Line 2: $\dfrac{x-1}{-2}=\dfrac{y-2}{-4}=\dfrac{z-3}{6}$, through $(1,2,3)$ with direction $(-2,-4,6)$.

Check parallel: is $(-2,-4,6)$ a scalar multiple of $(1,2,3)$? $-2/1=-2,\ -4/2=-2,\ 6/3=2$ — the third ratio disagrees, so the lines are not parallel. …

First rule out "parallel"/"coincident" by comparing direction ratios; then look for (or test for) a common point using the coplanarity determinant or, as here, by direct observa …

Stopping after checking the coplanarity determinant is zero and concluding "coplanar" without going the extra step to confirm the lines actually cross (as opposed to being parallel, which is also coplanar) — here the …