Write the line intersecting tfracx+52=(y-3)/(-4)=tfracz+13…

Write the line intersecting $\tfrac{x+5}2=\tfrac{y-3}{-4}=\tfrac{z+1}3$ and $\tfrac{x-3}{-2}=\tfrac{y+1}3=\tfrac{z+2}4$, and parallel to $3x+12y-3z-5=0$ and $3x-4y+9z+7=0$.

Explanation of Method

A line that meets two given lines and is parallel to two given planes.

Step 1: Direction

\[ \vec u=\{1,4,-1\}\times\{3,-4,9\}=\{32,-12,-16\}\parallel\{8,-3,-4\}. \]

Step 2: Intersection points

Take $A=(-5+2s,3-4s,-1+3s)$ and $B=(3-2t,-1+3t,-2+4t)$, with $\overrightarrow{AB}\parallel\{8,-3,-4\}$. This gives $13s+9t=4$ and $4s-3t=3$, so $s=\tfrac{13}{25}$ and $t=-\tfrac{23}{75}$:

\[ A\left(-\tfrac{99}{25},\tfrac{23}{25},\tfrac{14}{25}\right),\qquad B\left(\tfrac{271}{75},-\tfrac{144}{75},-\tfrac{242}{75}\right),\qquad \overrightarrow{AB}=\tfrac{71}{75}\{8,-3,-4\}. \]

\[\frac{x+\frac{99}{25}}{8}=\frac{y-\frac{23}{25}}{-3}=\frac{z-\frac{14}{25}}{-4}\]

$A$ $B$ $\ell_1$ $\ell_2$

The required line (teal) with direction $\{8,-3,-4\}$ meets $\ell_1$ at $A$ and $\ell_2$ at $B$.

Result

\[\boxed{\,\frac{x+\frac{99}{25}}{8}=\frac{y-\frac{23}{25}}{-3}=\frac{z-\frac{14}{25}}{-4}\,}\]

Remark

The source's first attempt has sign errors in (I) and (II), which led to $s=-13$. The source then corrects itself and obtains the result above, which has been verified.