Find the equation of the plane that passes through the point $$ P(1,2,3) $$ and has the normal vector $$ \mathbf{N}=[2,-1,4] $$.

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To find the equation of the plane that passes through the point $$ P(1,2,3) $$ and has the normal vector $$ \mathbf{N} = [2, -1, 4] $$, you can use the point-normal form of the equation of a plane.

The general form of the equation of a plane given a point $$ P(x_0, y_0, z_0) $$ and a normal vector $$ \mathbf{N} = [A, B, C] $$ is:

$$
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
$$

Plugging in the given values:

$$
2(x - 1) - 1(y - 2) + 4(z - 3) = 0
$$

Now, expand and simplify the equation:

$$
2x - 2 - y + 2 + 4z - 12 = 0
$$
$$
2x - y + 4z - 12 = 0
$$

This can be rewritten as:

$$
2x - y + 4z = 12
$$

**Final Equation of the Plane:**
$$
2x - y + 4z = 12
$$
The equation of the plane is $$ 2x - y + 4z = 12 $$.

Find the equation of the plane that passes through the point and has the normal vector .

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