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

$\frac{2}{7}$

The correct answer is option (2) → $\frac{2}{7}$

$\vec n=2\hat i-3\hat j+6\hat k$

plane makes $\sin^{-1}α$ with x axis

⇒ Normal makes $(90°-\sin^{-1}α)$ with x axis.

Let x axis be $\hat b=\hat i$

so $\vec n.\hat b=|\vec n||\hat b|\cos(90°-\sin^{-1}α)$

so $2=7\sin(\sin^{-1}α)⇒α=\frac{2}{7}$