The plane we want and the plane we're given are parallel, so they share the same normal vector. The normal to [tex]2x+3y-z=0[/tex] is the vector (2, 3, -1), since [tex](2,3,-1)\cdot(x,y,z)=0[/tex].
Then the plane we want has equation
[tex](2,3,-1) \cdot (x-3, y-2, z-5) = 0 \\\\ \implies 2(x-3) + 3(y-2) - (z-5) = 0 \\\\ \implies \boxed{2x + 3y - z = 7}[/tex]