The distance of the plane `6x-3y+2z-14=0` from the origin is

Options

A. 2
B. 1
C. 14
D. 8

Explanation

To find the distance of the plane \(6x - 3y + 2z - 14 = 0\) from the origin, we can use the formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane given by the equation \(Ax + By + Cz + D = 0\):

\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step 1: Identify the coefficients and the point
In the equation of the plane \(6x - 3y + 2z - 14 = 0\), we can identify:
- \(A = 6\)
- \(B = -3\)
- \(C = 2\)
- \(D = -14\)

The point we are considering is the origin, which is \((0, 0, 0)\).

### Step 2: Substitute the point into the distance formula
Substituting \(x_0 = 0\), \(y_0 = 0\), \(z_0 = 0\) into the formula gives:

\[ D = \frac{|6(0) + (-3)(0) + 2(0) - 14|}{\sqrt{6^2 + (-3)^2 + 2^2}} \]

### Step 3: Simplify the numerator
Calculating the numerator:

\[ D = \frac{|0 + 0 + 0 - 14|}{\sqrt{6^2 + (-3)^2 + 2^2}} = \frac{|-14|}{\sqrt{6^2 + 9 + 4}} = \frac{14}{\sqrt{49}} \]

### Step 4: Simplify the denominator
Calculating the denominator:

\[ \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]

### Step 5: Calculate the distance
Now substituting back into the distance formula:

\[ D = \frac{14}{7} = 2 \]

### Final Answer
Thus, the distance of the plane from the origin is \(2\) units.

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