The plane $2x-3y+6z-11=0$ makes an angle ${\sin }^{-1}(\alpha )$ with the X-axis. The value of $\alpha$ is equal to:

The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by:
$$\sin \theta =\frac{|\vec{n}\cdot \vec{b}|}{|\vec{n}||\vec{b}|}$$
Here, the normal to the plane is $\vec{n}=(2,-3,6)$.The direction vector of the X-axis is $\vec{b}=(1,0,0)$.
$$\sin \theta =\frac{|(2)(1)+(-3)(0)+(6)(0)|}{\sqrt{2^2+(-3{)}^2+6^2}\cdot \sqrt{1^2+0^2+0^2}}$$
$$\sin \theta =\frac{2}{\sqrt{4+9+36}\cdot 1}=\frac{2}{\sqrt{49}}=\frac{2}{7}$$
Given $\theta ={\sin }^{-1}(\alpha )$, we have $\alpha =2/7$.