Both planes have the same normal vector:
[tex]6x-y-z=6\implies\mathbf n=\langle6,-1,-1\rangle[/tex]
The plane we want must contain the point (4, -3, -1), so its equation would be
[tex]\mathbf n\cdot\langle x-4,y+3,z+1\rangle=0[/tex]
[tex]\implies 6(x-4)-(y+3)-(z+1)=0[/tex]
[tex]\implies\boxed{6x-y-z=28}[/tex]
