A vector ` vec n` of magnitude 8 units is inclined to x-axis at `45^0` , y-axis at `60^0` and an acute angle with z-axis. If a plane passes through a point `(sqrt(2),-1,1)` and is normal to ` vec n` , find its equation in vector form.

Options

A. `3sqrt(2)x-4y-3z=7`
B. `4sqrt(2)x+7y+z=2`
C. `sqrt(2)x+y+z=2`
D. `sqrt(2)x-y-z=2`

Explanation

To find the equation of the plane that passes through the point \((\sqrt{2}, -1, 1)\) and is normal to the vector \(\vec{n}\), we will follow these steps:

### Step 1: Determine the direction cosines
Given that the vector \(\vec{n}\) is inclined at \(45^\circ\) to the x-axis and \(60^\circ\) to the y-axis, we can find the direction cosines \(L\), \(M\), and \(N\):
- \(L = \cos(45^\circ) = \frac{1}{\sqrt{2}}\)
- \(M = \cos(60^\circ) = \frac{1}{2}\)

### Step 2: Use the relationship of direction cosines
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