Write the line that lies in the plane x-4y+2z-7=0 , passes…

Write the line that lies in the plane $x-4y+2z-7=0$, passes through the point where this plane meets the line $\ell$: $x-2y-4z+3=0$, $2x+y-3z+1=0$, and is perpendicular to $\ell$.

Explanation of Method

A line lying in a plane, through the piercing point of a given line, and perpendicular to it.

Step 1: Solution

\[ \vec d=\{1,-2,-4\}\times\{2,1,-3\}=\{10,-5,5\}\parallel\{2,-1,1\},\qquad Q(-1,1,0)\in\ell . \]

Substituting $x=-1+2t,\ y=1-t,\ z=t$ into the plane: $8t-12=0$, so $P\left(2,-\tfrac12,\tfrac32\right)$.

\[ \vec u=\vec n_\pi\times\vec d=\{1,-4,2\}\times\{2,-1,1\}=\{-2,3,7\}. \]

\[\frac{x-2}{-2}=\frac{y+\frac12}{3}=\frac{z-\frac32}{7}\]

Check: $\vec u\cdot\vec n_\pi=-2-12+14=0$ and $\vec u\cdot\vec d=-4-3+7=0$.

$P$ $\ell$ $m$

The line $\ell$ (red) pierces the plane at $P\left(2,-\tfrac12,\tfrac32\right)$; the required line $m$ (teal) lies in the plane and is perpendicular to $\ell$.

Result

\[\boxed{\,\frac{x-2}{-2}=\frac{y+\frac12}{3}=\frac{z-\frac32}{7}\,}\]