Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier, 
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$

My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\lambda(x^2+y^2+z^2-2x+2y+6z+9)=0$$
Differentiated partially wrt $x$, $y$ and $z$, and equated to $0$.
I get the following equations
$$(2\lambda+2)x-y-z=2\lambda$$
$$-x+(2\lambda+2)y-z=-2\lambda$$
$$-x-y+(2\lambda+2)z=6\lambda$$
In matrix form:
$$\begin{bmatrix} 2\lambda+2 & -1 & -1 \\-1 & 2\lambda+2 & -1 \\ -1 & -1 & 2\lambda+2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}2 \\ -2 \\ -6 \end{bmatrix} $$
When I try to find $x$, $y$ and $z$ from these equations, it gets complex. Is there an easier way to ascertain their values?

You already got the system of equations. So the solution simply follows from there.
You should get $(x,y,z)=(2,-1,-4)$ and $(x,y,z)=(0,-1,-2)$ which gives you the maximum (27) and the minimum (3), respectively. 
