The distance of the point (1, -2, 4) from the plane passing through the point (1,2, 2) and perpendicular to the planes x - y + 2z = 3 and 2x- 2y + z + 12 = 0, is

$2\sqrt{2}$

Let $\vec{n}$ be a vector normal to the required plane.

Then, its equation is 

$\left(\vec{r}- (\hat{i} + 2\hat{j} + 2\hat{k})\right). \vec{n} = 0 $ ......(i)

This plane is perpendicular to the planes x - y + 2z = 3 and 2x - 2y + z + 12 = 0 whose normals $\vec{n_1}= \hat{i} - \hat{j}+2\hat{k} $ and $\vec{n_2}= 2\hat{i} - 2 \hat{j} + \hat{k}.$

$∴ \vec{n} = \vec{n_1}× \vec{n_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -1 & 2\\2 & -2 & 1\end{vmatrix}=3\hat{i}+3\hat{j}+0\hat{k}$

Substituting $\vec{n} = 3\hat{i} + 3\hat{j}$ in (i), we obtain

$\vec{r}. ( 3\hat{i} + 3\hat{j})= 9 $ or, $\vec{r}. ( \hat{i} + \hat{j})= 3$ ....(ii)

The distance of this plane from $\vec{a} = \hat{i} - 2\hat{j} + 4\hat{k}$ is given by

$ d = \begin{vmatrix}\frac{(\hat{i}-2\hat{j}+4\hat{k}).(\hat{i}+\hat{j})-3}{\sqrt{1+1}}\end{vmatrix}= 2\sqrt{2}$