The equation of the plane through the point (1,2,3) and parallel to the plane `x+2y+5z=0` is

Options

A. `(x-1)+2(y-2)+5(z-3)=0`
B. `x+2y+5z=14`
C. `x+2y+5z=6`
D. none of these

Explanation

To find the equation of the plane that passes through the point (1, 2, 3) and is parallel to the plane given by the equation \(x + 2y + 5z = 0\), we can follow these steps:

### Step 1: Identify the normal vector of the given plane
The normal vector of the plane \(x + 2y + 5z = 0\) can be derived from the coefficients of \(x\), \(y\), and \(z\). Thus, the normal vector \( \mathbf{n} \) is:
\[ \mathbf{n} = (1, 2, 5) \]

### Step 2: Use the point-normal form of the plane equation
The general equation of a plane in point-normal form is given by:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
where \((x_1, y_1, z_1)\) is a point on the plane and \((a, b, c)\) are the components of the normal vector.

### Step 3: Substitute the known values
Here, the point through which the plane passes is \((1, 2, 3)\) and the normal vector is \((1, 2, 5)\). Substituting these values into the equation gives:
\[ 1(x - 1) + 2(y - 2) + 5(z - 3) = 0 \]

### Step 4: Expand the equation
Now, we will expand the equation:
\[ x - 1 + 2y - 4 + 5z - 15 = 0 \]
Combining like terms, we get:
\[ x + 2y + 5z - 20 = 0 \]

### Step 5: Rearrange the equation
Rearranging the equation, we can write it in standard form:
\[ x + 2y + 5z = 20 \]

Thus, the equation of the plane through the point (1, 2, 3) and parallel to the plane \(x + 2y + 5z = 0\) is:
\[ \boxed{x + 2y + 5z = 20} \]

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