If $x^2+9 y^2+4 z^2=12(x-2 y+2 z)-88$, then the value of $(x-3 y+z)$ is:

13

If $x^2+9 y^2+4 z^2=12(x-2 y+2 z)-88$

Make the above equation in the form of 

a² + b² + c²  = (a + b + c)² - 2(ab + bc + ac)

$x^2+9 y^2+4 z^2= 2(6x- 12y+12z)-88$

we can find the values of the variables by =

Coefficient of variables on right sides divide by coefficient of same variable on left side along with the signs as given below = 

x = 6

4 = -\(\frac{12}{9}\)

c = \(\frac{12}{4}\) = 3

$(x-3 y+z)$ = (6 -3 × (-\(\frac{12}{9}\)) + 3) = 13