A vector \(\vec{n}\) is inclined to \(X\)-axis at \(45^{\circ}\), \(\mathrm{Y}\)-axis at \(60^{\circ}\) and at an acute angle to \(\mathrm{Z}\)-axis. If \(\overrightarrow{\mathrm{n}}\) is normal to a plane passing through the point \((-\sqrt{2}, 1,1)\), then equation of the plane is

Let \(\vec{n}\) be inclined at angles \(\alpha, \beta, \gamma\) to \(\mathrm{X}, \mathrm{Y}, \mathrm{Z}\) axes respectively.
\(\alpha=45^{\circ}, \beta=60^{\circ}, \gamma=? \)
\( \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \)
\( \therefore \frac{1}{2}+\frac{1}{4}+\cos ^2 \gamma=1 \)
\( \therefore \cos ^2 \gamma=\frac{1}{4} \)
\( \therefore \gamma=60^{\circ} \)
\( \overrightarrow{\mathrm{n}}=\cos \alpha \hat{\mathrm{i}}+\cos \beta \hat{\mathrm{j}}+\cos \gamma \hat{\mathrm{k}} \)
\( \therefore \overrightarrow{\mathrm{n}}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\)
\(\therefore\) Equation of the required plane is
\(\frac{1}{\sqrt{2}}(x+\sqrt{2})+\frac{1}{2}(y-1)+\frac{1}{2}(z-1)=0\)
i.e., \(\sqrt{2} x+y+z=0\)