If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is    what?    
$108$ , $216$ , $405$ , $1048$

Let $F=1/x+1/y+1/z-1$ and $G=x^2+8y^2+27z^2$. You seek min of $G$ with constraint $F=0.$  Scalar multiples of the gradients are 
$$-\nabla F=(1/x^2,1/y^2,1/z^2),$$
$$(1/2)\nabla G=(x,8y,27z).$$
Using Lagrange multipliers this means at a critical point the following are zero:
$$F_xG_y-F_yG_x=-(x-2y)(x^2+2xy+4y^2),$$
$$F_yG_z-G_yF_z=-(2y-3z)(4y^2+6yz+9z^2),$$
$$F_xG_z-G_xF_z=-(x-3z)(x^2+3xz+9z^2).$$
Since the variables are positive this means at a critical point $x=2y,\ 2y=3z,\ x=3z$ which with the constraint $F=0$ gives the single critical point $(6,3,2)$. 
At $(6,3,2)$ we have $G=216.$ This is one of the choices in the question. Note that at $(2,3,6)$ the value is $G=1048,$ one of the other choices, leading one to conjecture the test makers switched the coordinates to come up with another multiple choice possibility.
NOTE: I haven't rigorously proved that $G$ takes on its absolute minimum at the critical point $(6,3,2)$. But it seems likely, in that moving one of the variables near $0$ causes one of the others to get large, making $G$ become large since the variables are squared in $G$. If I can rigorize I'll add more to this answer. Or if someone else can do so that would be good.
