Point $C$ is

$\left(\frac{2k-3}{k+1},\frac{4k-6}{k+1},\frac{-3k+1}{k+1}\right)$

$\frac{x-1}{-1}=\frac{y+4}{2}=\frac{z+2}{3}$

Plane $lx+my+nz=0$

$l\left(-1\right)+m\left(2\right)+n\left(3\right)=0$

$-l+2m+3n=0$   $...\left(1\right)$

It also satisfy point $\left(1,-4,-2\right)$

$l-4m-2n=0$   $...\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$2m+3n=4m+2n$

$n=2m$

$l-4m-4m=0$

$l=8 \mathrm{m}$

$\frac{l}{8}=\frac{m}{1}=\frac{n}{2}$

$l:m:n=8:1:2$

Plane is $8x+y+2z=0$

It will satisfy point $C$

$8\left(\frac{2k-3}{k+1}\right)+\left(\frac{4k-6}{k+1}\right)+2\left(\frac{-3k+1}{k+1}\right)=0$

$16k-24+4k-6-6k+2=0$

$14k=28 \therefore k=2$