The distance of plane $\vec{r}.(6\hat{i}-3\hat{j}-2\hat{k}) + 1 = 0$ from origin is : 

$\frac{1}{7}$

ax + by + cz + d = 0

Distance = $\frac{|d|}{\sqrt{a^2+b^2+c^2}}$

$(x\hat i+y\hat j+z\hat k)(6\hat i-3\hat j-2\hat k)+10=0$

Distance = $\frac{|-1|}{\sqrt{(6)^2+(-3)^2+(-2)^2}}=\frac{1}{\sqrt{36+9+4}}=\frac{1}{\sqrt{49}}$

$=\frac{1}{7}$