The distance of a point (1, -1, 0) from the plane $\vec{r}. (3\hat{i}-2\hat{j}-6\hat{k})=4$ is :

$\frac{1}{7}$

Plane: $3x-2y-6z=4 \;\;( \text{normal } \vec{n}=\langle 3,-2,-6\rangle)$, Point: $P(1,-1,0)$

Distance formula: $\displaystyle d=\frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$ for $ax+by+cz=d$

Here $a=3,\;b=-2,\;c=-6,\;d=4$

Numerator $=\;|3\cdot1+(-2)\cdot(-1)+(-6)\cdot0-4|\;=\;|3+2+0-4|\;=\;|1|\;=\;1$

Denominator $=\;\sqrt{3^2+(-2)^2+(-6)^2}\;=\;\sqrt{9+4+36}\;=\;\sqrt{49}\;=\;7$

Distance: $\displaystyle d=\frac{1}{7}$