The distance of the plane through (1, 1, 1) and perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ from the origin is

$\frac{7}{5}$

The given line is normal to the plane. SO, direction ratios of normal to the plane are proportional to 3, 0, 4. The plane passes through (1, 1, 1).SO, its equation is

$3(x-1) + 0 (y-1) + 4(z-1) = 0 $ or, $ 3x + 4z - 7 = 0 $

$d=\frac{|-7|}{\sqrt{3^2 + 0^2 + 4^2}}=\frac{7}{5}$