If the plane `2ax-3ay+4az+6=0` passes through the mid point of the line joining the centre of the spheres `x^(2)+y^(2)+z^(2)+6x-8y-2z=13 and x^(2)+y^(2)+z^(2)-10x+4y-2z=8`, then `alpha` equals

Options

A. `2`
B. `-2`
C. `1`
D. `-1`

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript:

### Step 1: Identify the equations of the spheres
The equations of the spheres given are:
1. \( x^2 + y^2 + z^2 + 6x - 8y - 2z = 13 \)
2. \( x^2 + y^2 + z^2 - 10x + 4y - 2z = 8 \)

### Step 2: Convert the equations of the spheres to standard form
To convert the equations into standard form, we complete the square for each variable.

**For the first sphere:**
\[ x^2 + 6x + y^2 - 8y + z^2 - 2z = 13 \]
Completing the square:
- For \(x\): \(x^2 + 6x = (x + 3)^2 - 9\)
- For \(y\): \(y^2 - 8y = (y - 4)^2 - 16\)
- For \(z\): \(z^2 - 2z = (z - 1)^2 - 1\)

Substituting back:
\[ (x + 3)^2 - 9 + (y - 4)^2 - 16 + (z - 1)^2 - 1 = 13 \]
\[ (x + 3)^2 + (y - 4)^2 + (z - 1)^2 = 39 \]

**For the second sphere:**
\[ x^2 - 10x + y^2 + 4y + z^2 - 2z = 8 \]
Completing the square:
- For \(x\): \(x^2 - 10x = (x - 5)^2 - 25\)
- For \(y\): \(y^2 + 4y = (y + 2)^2 - 4\)
- For \(z\): \(z^2 - 2z = (z - 1)^2 - 1\)

Substituting back:
\[ (x - 5)^2 - 25 + (y + 2)^2 - 4 + (z - 1)^2 - 1 = 8 \]
\[ (x - 5)^2 + (y + 2)^2 + (z - 1)^2 = 38 \]

### Step 3: Find the centers of the spheres
The centers of the spheres are:
1. Center of the first sphere: \((-3, 4, 1)\)
2. Center of the second sphere: \((5, -2, 1)\)

### Step 4: Calculate the midpoint of the line joining the centers
The midpoint \(M\) of the line joining the centers is calculated as follows:
\[ M = \left( \frac{-3 + 5}{2}, \frac{4 - 2}{2}, \frac{1 + 1}{2} \right) = \left( \frac{2}{2}, \frac{2}{2}, \frac{2}{2} \right) = (1, 1, 1) \]

### Step 5: Substitute the midpoint into the plane equation
The equation of the plane is given by:
\[ 2ax - 3ay + 4az + 6 = 0 \]
Substituting the midpoint \((1, 1, 1)\):
\[ 2a(1) - 3a(1) + 4a(1) + 6 = 0 \]
This simplifies to:
\[ 2a - 3a + 4a + 6 = 0 \]
\[ 3a + 6 = 0 \]
\[ 3a = -6 \]
\[ a = -2 \]

### Final Answer
Thus, the value of \(a\) is \(-2\).