Explanation:The plane is given as, \(\overrightarrow {\rm{r}} \,{\rm{.}}\,\left( {{\rm{2\hat i}}\,{\rm{ - }}\,{\rm{\hat j}}\,{\rm{ + }}\,{\rm{2\hat k}}} \right)=12\).We know that \(\vec r=x\hat{i}+y\hat{j}+z\hat{k}\)We can write the equation of the plane in general form as,2x - y + 2z - 12 = 0 ...... (1)Now, to get the normal form of a plane given in general form as,Ax + By + Cz + D = O where D ? 0......(2),Distance between plane and origin is given as:\(p=-\frac{-D}{|\vec n|}\)we have to divide the equation (1) by \(|\vec {n}|\) , where \(\vec {n}\) is the normal vector given as, \(\vec{n}=\left( {{\rm{2\hat i}}\,{\rm{ - }}\,{\rm{\hat j}}\,{\rm{ + }}\,{\rm{2\hat k}}} \right)\)Now, \(|\vec{n}|=\left( {{\rm{2\hat i}}\,{\rm{ - }}\,{\rm{\hat j}}\,{\rm{ + }}\,{\rm{2\hat k}}} \right)\)\(|\vec n|=\sqrt{2^2+(-1)^2+(2)^2}=3\)Comparing equation (1) with equation (2), we get, D = -12\(p=-\frac{-D}{|\vec n|}\)\(p=-\frac{-12}{3}=4\)