If the system of linear equations given by x + y + z = 3 ,   2 x + 2 y - z = 3 ,   x + y - z = 1…

Given,

$x+y+z=3 ...\left(i\right)$

$2x+2y-z=3 ...\left(ii\right)$

$x+y-z=1 ...\left(iii\right)$

Adding, we get

$x+y=2 ...\left(iv\right)$

Substituting this value of $\left(x+y\right)$ in $\left(i\right)$, we get

$z=1$

Now, if $\left(x_0,y_0,z_0\right)$ is the solution then $x_0+y_0=2 \& z_0=1$.

Then,

$2x_0+2y_0+z_0=2\left(x_0+y_0\right)+z_0$

$=2\times 2+1=5$.