If the equation of the plane passing through `(1,2, 3)` and situated at a maximum distance from point `(2, 3, 4)` is P = 0, then the distance (in units) of P = 0 from origin is

Options

A. `sqrt3`
B. `2sqrt3`
C. `sqrt6`
D. `3sqrt2`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript and derive the equation of the plane and its distance from the origin.

### Step 1: Identify the points
We have two points:
- Point A (1, 2, 3) through which the plane passes.
- Point B (2, 3, 4) from which the plane is at maximum distance.

### Step 2: Find the direction vector
The direction vector from point A to point B is given by:
\[ \vec{AB} = B - A = (2 - 1, 3 - 2, 4 - 3) = (1, 1, 1) \]

### Step 3: Determine the normal vector of the plane
Since the plane is at maximum distance from point B, the normal vector of the plane must be parallel to the vector \(\vec{AB}\). Therefore, the normal vector \(\vec{n}\) can be taken as:
\[ \vec{n} = (1, 1, 1) \]

### Step 4: Write the equation of the plane
The general equation of a plane can be written as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((n_1, n_2, n_3)\) are the components of the normal vector.

Substituting point A (1, 2, 3) and the normal vector (1, 1, 1):
\[ 1(x - 1) + 1(y - 2) + 1(z - 3) = 0 \]
This simplifies to:
\[ x - 1 + y - 2 + z - 3 = 0 \]
or
\[ x + y + z - 6 = 0 \]
Thus, the equation of the plane is:
\[ P: x + y + z = 6 \]

### Step 5: Calculate the distance from the origin to the plane
The distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane \(x + y + z - 6 = 0\), we have:
- \(A = 1\), \(B = 1\), \(C = 1\), and \(D = -6\)
- The point is the origin \((0, 0, 0)\)

Substituting these values into the distance formula:
\[ d = \frac{|1(0) + 1(0) + 1(0) - 6|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|0 - 6|}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

### Final Answer
The distance of the plane \(P = 0\) from the origin is \(2\sqrt{3}\) units.

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