The length of perpendicular from the point (1, 0, 1) to the plane x - y + z = 4 is:

$\frac{2}{\sqrt{3}}$

The correct answer is Option (1) → $\frac{2}{\sqrt{3}}$

$\vec n=\hat i-\hat j+\hat k$

let a point be fool of perpendicular on planes from (1, 0, 1)

so eq. of line along $\vec n$ containing (1, 0, 1)

$\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{1}=λ$

$x=λ+1,y=-λ,z=λ+1$

so $λ+1+λ+λ+1=4$ (in eq. of plane)

so $3λ=2$

$λ=\frac{2}{3}$

$x=\frac{5}{3},y=-\frac{2}{3},z=\frac{5}{3}$

So distance = $\sqrt{(1-\frac{5}{3})^2+(0+\frac{2}{3})^2+(1-\frac{5}{3})^2}$

$=\frac{2}{\sqrt{3}}$