1.A vector n⃗\vec{n}n is inclined to x-axis at 45∘45^\circ45∘, to y-axis at 60∘60^\circ60∘ and at an acute angle to z-axis. If n⃗\vec{n}n is a normal to a plane passing through the point (2,−1,1)(\sqrt{2}, -1, 1)(2,−1,1), then the equation of the plane isa.42x+7y+z=24\sqrt{2}x + 7y + z = 242x+7y+z=2b.2x+y+2z=22+12x + y + 2z = 2\sqrt{2} + 12x+y+2z=22+1c.32x−4y−3z=73\sqrt{2}x - 4y - 3z = 732x−4y−3z=7d.2xyz=2\sqrt{2}xyz = 22xyz=2Login to continueOnly logged in users canattempt or see the solution.