Write the line through M_1(3,-2,-4) parallel to 3x-2y-3z-7=0…

Write the line through $M_1(3,-2,-4)$ parallel to $3x-2y-3z-7=0$ and intersecting $\tfrac{x-2}3=\tfrac{y+4}{-2}=\tfrac{z-1}2$.

Explanation of Method

The line through M₁ that is parallel to a plane and meets a given line.

Step 1: Solution

Take $A=(2+3t,-4-2t,1+2t)$. Then $\overrightarrow{M_1A}=\{3t-1,-2-2t,5+2t\}$, and $\overrightarrow{M_1A}\cdot\{3,-2,-3\}=7t-14=0$ gives $t=2$, so $A(8,-8,5)$.

\[\frac{x-3}{5}=\frac{y+2}{-6}=\frac{z+4}{9}\]

Check: $15+12-27=0$, and $M_1$ is not in the plane ($18\ne0$).

$M_1$ $A(8,-8,5)$ $\ell$

The required line (red) through $M_1$ meets $\ell$ at $A$. It lies in the plane through $M_1$ parallel to $3x-2y-3z-7=0$ (gray).

Result

\[\boxed{\,\frac{x-3}{5}=\frac{y+2}{-6}=\frac{z+4}{9}\,}\]