Determine n so that (x-1)/(1)=(y-2)/(-1)=(z-1)/(n)…

Determine $n$ so that $\tfrac{x-1}{1}=\tfrac{y-2}{-1}=\tfrac{z-1}{n}$ intersects $\tfrac{x-2}{1}=\tfrac{y-3}{2}=\tfrac{z-4}{3}$, and find the intersection point.

Explanation of Method

Finding n so that two lines intersect, and their intersection point.

Step 1: Coplanarity

\[ \begin{vmatrix}-1&-1&-3\\1&-1&n\\1&2&3\end{vmatrix}=(3+2n)+(3-n)-9=n-3=0\ \Longrightarrow\ n=3 . \]

The directions $\{1,-1,3\}$ and $\{1,2,3\}$ are not parallel, so coplanar here means intersecting.

Step 2: Intersection point

Set $1+s=2+t$ and $2-s=3+2t$. Then $t=-\tfrac23$, $s=\tfrac13$, and the $z$-values agree: $1+1=4-2=2$.

\[n=3,\qquad P\left(\tfrac43,\tfrac53,2\right)\]

$M_1$ $M_2$ $P\left(\tfrac43,\tfrac53,2\right)$

$\ell_1\ (n=3)$
$\ell_2$

For $n=3$ the two lines meet at $P\left(\tfrac43,\tfrac53,2\right)$.

Result

\[\boxed{\,n=3,\qquad P\left(\tfrac43,\tfrac53,2\right)\,}\]