The shortest distance from the plane `12 x+y+3z=327`to the sphere `x^2+y^2+z^2+4x-2y-6z=155` is

Options

A. 39
B. 26
C. `41(4)/(13)`
D. 13

Explanation

To find the shortest distance from the plane \(12x + y + 3z = 327\) to the sphere \(x^2 + y^2 + z^2 + 4x - 2y - 6z = 155\), we will follow these steps:

### Step 1: Rewrite the equation of the sphere in standard form

The equation of the sphere is given as:
\[ x^2 + y^2 + z^2 + 4x - 2y - 6z = 155 \]
We need to complete the square for \(x\), \(y\), and \(z\).

1. For \(x^2 + 4x\):
\[ x^2 + 4x = (x + 2)^2 - 4 \]

2. For \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

3. For \(z^2 - 6z\):
\[ z^2 - 6z = (z - 3)^2 - 9 \]

Now substituting these back into the sphere equation:
\[ (x + 2)^2 - 4 + (y - 1)^2 - 1 + (z - 3)^2 - 9 = 155 \]
This simplifies to:
\[ (x + 2)^2 + (y - 1)^2 + (z - 3)^2 - 14 = 155 \]
Thus,
\[ (x + 2)^2 + (y - 1)^2 + (z - 3)^2 = 169 \]

### Step 2: Identify the center and radius of the sphere

From the equation \((x + 2)^2 + (y - 1)^2 + (z - 3)^2 = 169\):
- The center of the sphere \(C\) is \((-2, 1, 3)\).
- The radius \(r\) is \(\sqrt{169} = 13\).

### Step 3: Calculate the perpendicular distance from the center of the sphere to the plane

The formula for the perpendicular distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For our plane \(12x + y + 3z - 327 = 0\):
- \(A = 12\), \(B = 1\), \(C = 3\), and \(D = -327\).
- The center of the sphere is \((-2, 1, 3)\).

Substituting these values into the distance formula:
\[ D = \frac{|12(-2) + 1(1) + 3(3) - 327|}{\sqrt{12^2 + 1^2 + 3^2}} \]
Calculating the numerator:
\[ = | -24 + 1 + 9 - 327 | = |-341| = 341 \]

Calculating the denominator:
\[ \sqrt{12^2 + 1^2 + 3^2} = \sqrt{144 + 1 + 9} = \sqrt{154} \]

Thus, the distance \(D\) becomes:
\[ D = \frac{341}{\sqrt{154}} \]

### Step 4: Determine the shortest distance from the sphere to the plane

The shortest distance from the sphere to the plane is given by:
\[ \text{Shortest Distance} = D - r \]
Substituting the values we found:
\[ \text{Shortest Distance} = \frac{341}{\sqrt{154}} - 13 \]

### Final Answer

Thus, the shortest distance from the plane to the sphere is:
\[ \text{Shortest Distance} = \frac{341}{\sqrt{154}} - 13 \]

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