A plane makes positive intercepts of unit length on each of \(X\) and \(Y\) axis. If it passes through the point \((-1,1,2)\) and makes angle \(\theta\) with the X -axis, then \(\theta\) is

Equation of plane in intercept form is
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
Here, \(\mathrm{a}=1, \mathrm{~b}=1\)
\(\therefore \quad \frac{x}{1}+\frac{y}{1}+\frac{z}{c}=1\)
Since this plane passes through the point \((-1,1,2)\)
\(\begin{aligned}
\therefore \quad & -1+1+\frac{2}{c}=1 \\
& \Rightarrow c=2
\end{aligned}\)
\(\therefore \quad\) Equation of plane is
\(\begin{aligned}
& x+y+\frac{z}{2}=1 \\
& \Rightarrow 2 x+2 y+z=2
\end{aligned}\)
D.r.s of X -axis are \(1,0,0\).
\(\begin{aligned}
& \therefore \quad \sin \theta=\frac{2(1)+0+0}{\sqrt{4+4+1} \sqrt{1}} \\
& \quad \Rightarrow \sin \theta=\frac{2}{3} \\
& \quad \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{3}\right)
\end{aligned}\)