A vector $$\vec{n}$$ is inclined to the $$x$$-axis at 45°, to the $$y$$-axis at 60° and at an acute angle to the $$z$$-axis. If $$\vec{n}$$ is a normal to a plane passing through the point $$(\sqrt{2}, -1, 1)$$ then the equation of the plane is :

The orientation of a vector in 3D space is defined by its direction cosines $$(l, m, n)$$, where $$l = \cos \alpha$$, $$m = \cos \beta$$, and $$n = \cos \gamma$$. These satisfy the identity: $$l^2 + m^2 + n^2 = 1$$

Angle with $$x$$-axis, $$\alpha = 45^\circ \implies l = \cos 45^\circ = \frac{1}{\sqrt{2}}$$, and Angle with $$y$$-axis, $$\beta = 60^\circ \implies m = \cos 60^\circ = \frac{1}{2}$$

$$\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1$$

Since $$\gamma$$ is acute, $$n = \cos \gamma = \frac{1}{2}$$

Thus, the direction cosines of the normal vector are $$(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2})$$ or $$(a, b, c) = (\sqrt{2}, 1, 1)$$

Equation of plane:

$$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$

$$\sqrt{2}(x - \sqrt{2}) + 1(y - (-1)) + 1(z - 1) = 0$$

$$\sqrt{2}x + y + z = 2$$