$\text{ Projection of the line }\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}\text{ on the plane }x+2y+z=9\text{ is the line of intersection of plane }x+2y+z=9$ $\text{with the plane}$

$\text{ Projection of given line }L\equiv \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}\text{ on }$

$\text{ given plane }P\equiv x+2y+z=9\text{ is the line of intersection of plane }$

$P\text{ and plane }\left(ABCD\right)\text{ through the line }L\text{ and perpendicular to the plane P.}$

$\text{ Now line }L\text{ is parallel to vector }\overrightarrow{a}=2\widehat{i}-\widehat{j}+4\widehat{k}----\left(1\right)$

$\text{ Vector normal to plane }P\text{ is }\overrightarrow{n}=\widehat{i}+2\widehat{j}+\widehat{k}----\left(2\right)$

$\therefore \text{ Vector normal plane }ABCD\text{ is }\overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{n}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& -1& 4\\ 1& 2& 1\end{array}\right|$

$=-9\widehat{i}+2\widehat{j}+5\widehat{k}$

Plane ABCD passes through the line L, i.e., passes through the
point (1, -1, 3). Thus,

$\text{ Equation of plane }ABCD=(-9)(x-1)+2(y+1)+5(z-3)=0$

$\text{ or }9x-2y-5z+4=0----\left(3\right)$

Now required projection is line of intersection of given plane P and plane ABCD

$or x+2y+z-9=9x-2y-5z+4$