The distance of the point P(–1, –2, 4) from the plane x + y + z = 7 measured parallel to z-axis is

Equation of line parallel to z-axis passing through A is   $\frac{x+1}{0}=\frac{y+2}{0}=\frac{z-4}{1}=\lambda$
Point on the line is  $(-1,-2,\lambda +4)$ which lies on x + y + z = 7  
$\Rightarrow \lambda =6\&Q=(-1,-2,10)$ So, required distance is 6 units