The plane we want to find has general equation
[tex]a(x-2)+b(y-5)+c(z-7)=0[/tex]
with [tex]a,b,c[/tex] not equal to 0, and has normal vector
[tex]\vec n=a\,\vec\imath+b\,\vec\jmath+c\,\vec k[/tex]
[tex]\vec n[/tex] is perpendicular to both the normal vector of the other plane, which is [tex]4\,\vec\imath+5\,\vec\jmath+6\,\vec k[/tex], as well as the tangent vector to the line [tex]\vec r(t)[/tex], which is [tex]\vec\imath+2\,\vec\jmath+9\,\vec k[/tex].
This means the dot product of [tex]\vec n[/tex] with either vector is 0, giving us
[tex]\begin{cases}4a+5b+6c=0\\a+2b+9c=0\end{cases}[/tex]
Suppose we fix [tex]c=1[/tex]. Then the system reduces to
[tex]\begin{cases}4a+5b=-6\\a+2b=-9\end{cases}[/tex]
and we get
[tex](4a+5b)-4(a+2b)=-6-4(-9)\implies-3b=30\implies b=-10[/tex]
[tex]a+2(-10)=-9\implies a=11[/tex]
Then one equation for the plane could be
[tex]\boxed{11(x-2)-10(y-5)+(z-7)=0}[/tex]
or in standard form,
[tex]\boxed{11x-10y+z=-21}[/tex]
The solution is unique up to non-zero scalar multiplication, which is to say that any equation [tex](11x-10y+z)k=-21k[/tex] would be a valid answer. For example, suppose we instead let [tex]c=2[/tex]; then we would have found [tex]a=22[/tex] and [tex]b=-20[/tex], but clearly dividing both sides of the equation
[tex]22(x-2)-20(y-5)+2(z-7)=0[/tex]
by 2 gives the same equation as before.
