If 4x
2
+y
2
+z
2
+41=2(4x+y+6z), then the value of 2x+y−z is:

Let's rearrange the given equation by moving all terms to the left-hand side:
4x
2
+y
2
+z
2
+41−8x−2y−12z=0
Now, group the terms corresponding to each variable to complete perfect squares:

(4x
2
−8x)+(y
2
−2y)+(z
2
−12z)+41=0
[(2x)
2
−2(2x)(2)+2
2
]−4+[y
2
−2(y)(1)+1
2
]−1+[z
2
−2(z)(6)+6
2
]−36+41=0
(2x−2)
2
+(y−1)
2
+(z−6)
2
−4−1−36+41=0
(2x−2)
2
+(y−1)
2
+(z−6)
2
−41+41=0
(2x−2)
2
+(y−1)
2
+(z−6)
2
=0
Since the sum of squares of real numbers is zero, each individual squared term must equal zero:

2x−2=0⟹2x=2⟹x=1

y−1=0⟹y=1

z - 6 = 0 \implies z = 6$

Now, find the value of 2x+y−z:

2x+y−z=2(1)+1−6=2+1−6=−3