A vector $\vec{n}$ is inclined to X-axis at 45°, Y-axis at 60° and at an acute angle to Z-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2},1,1)$ then equation of the plane is
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Logic Tip: As soon as you discover the normal vector's direction ratios are $(\sqrt{2}, 1, 1)$, check the options. Option A is the only one with this sequence of coefficients! You don't even need to plug in the point.