The line is given by vec r×(2vec i+vec j-3vec k)=5vec i+2vec…

The line is given by $\vec r\times(2\vec i+\vec j-3\vec k)=5\vec i+2\vec j+4\vec k$. Write its canonical equations.

Step 1: Meaning

$\vec r\times\vec d=\vec c$ (with $\vec c\perp\vec d$) is the line with direction $\vec d$ through any $\vec r_0$ satisfying $\vec r_0\times\vec d=\vec c$. Here $\vec d=\{2,1,-3\}$, $\vec c=\{5,2,4\}$, and $\vec c\cdot\vec d=10+2-12=0$. $\checkmark$

Step 2: A point

\[ \vec r_0\times\vec d=\{-3y_0-z_0,\ 3x_0+2z_0,\ x_0-2y_0\}=\{5,2,4\}. \]

With $z_0=0$: $x_0=\tfrac23$ and $y_0=-\tfrac53$. The third equation checks: $\tfrac23+\tfrac{10}3=4$.

\[\frac{x-\frac23}{2}=\frac{y+\frac53}{1}=\frac{z}{-3}\quad\Longleftrightarrow\quad\frac{3x-2}{6}=\frac{3y+5}{3}=\frac{z}{-3}\]