Write the line lying in 2x+3y-z+1=0 , making a minimal angle…

Write the line lying in $2x+3y-z+1=0$, making a minimal angle with $\tfrac{x-3}1=\tfrac{y-3}2=\tfrac z1$, and at minimal distance from the origin.

Explanation of Method

Among the lines of minimal angle, the one closest to the origin.

Step 1: Solution

The direction is $\{0,1,3\}$, as in Problem 276. Parametrise the lines by $x=a$: through $P(a,y_0,2a+3y_0+1)$, \[d^2=\frac{|\overrightarrow{OP}\times\vec u|^2}{|\vec u|^2}=\frac{14a^2+4a+1}{10}\ \Longrightarrow\ a=-\tfrac17,\quad d_{\min}=\frac{1}{\sqrt{14}} .\]

\[x=-\frac17,\qquad \frac{y}{1}=\frac{z-\frac57}{3}\]

Geometric view: $d_{\min}=\tfrac1{\sqrt{14}}$ is exactly the distance from $O$ to the plane. So the optimal line passes through the foot $N=-\tfrac1{14}\{2,3,-1\}=\left(-\tfrac17,-\tfrac3{14},\tfrac1{14}\right)$, which indeed lies on it ($t=-\tfrac3{14}$).

$M(0,0,1)$ $N$ $O$

$276:\ x=0,\ \tfrac y1=\tfrac{z-1}3$
$277:\ x=-\tfrac17,\ \tfrac y1=\tfrac{z-5/7}{3}$

Both lines have the direction $\{0,1,3\}$ (projection of $\{1,2,1\}$ onto the plane). The one of Problem 277 passes through $N$, the foot of the perpendicular from $O$ to the plane, so its distance from $O$ is $\tfrac1{\sqrt{14}}$.

Result

\[\boxed{\,x=-\frac17,\qquad \frac{y}{1}=\frac{z-\frac57}{3}\,}\]