Given equation of the plane is 

x + y + z = 1

1x + 1y + 1z = 1

Comparing with ax + by + cz =d

a = 1, b = 1, c = 1, d = 1

$\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}=\sqrt{1^2+1^2+1^2}=\sqrt{3}$

Direction cosincs of the normal to the plane are

$\mathrm{\ell }=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\mathrm{m}=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\mathrm{n}=\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$

$\mathrm{\ell }=\frac{1}{\sqrt{3}},\mathrm{m}=\frac{1}{\sqrt{3}},\mathrm{n}=\frac{1}{\sqrt{3}}$

∴   Direction consine of the normal to the plane are 

$=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$

Disatnce from the original $=\frac{\mathrm{d}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}=\frac{1}{\sqrt{3}}$