Statement 1: let `A( vec i+ vec j+ vec k)a n dB( vec i- vec j+ vec k)` be two points. Then point `P(2 vec i+3 vec j+ vec k)` lies exterior to the sphere with `A B` as its diameter. ,
Statement 2: If `Aa n dB` are any two points and `P` is a point in space such that ` vec(PA).vec(PB) >0` , then point `P` lies exterior to the sphere with `A B` as its diameter.

Options

A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.
C. Statement 1 is true and Statement 2 is false.
D. Statement 1 is false and Statement 2 is true.

Explanation

To determine whether the point \( P(2 \vec{i} + 3 \vec{j} + \vec{k}) \) lies outside the sphere with \( AB \) as its diameter, we will follow these steps:

### Step 1: Identify Points A and B
The points are given as:
- \( A = \vec{i} + \vec{j} + \vec{k} \)
- \( B = \vec{i} - \vec{j} + \vec{k} \)

### Step 2: Calculate the Vector \( \vec{AB} \)
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (\vec{i} - \vec{j} + \vec{k}) - (\vec{i} + \vec{j} + \vec{k}) = -2\vec{j} \]

### Step 3: Find the Midpoint M of AB
The midpoint \( M \) of the segment \( AB \) is given by:
\[ M = \frac{A + B}{2} = \frac{(\vec{i} + \vec{j} + \vec{k}) + (\vec{i} - \vec{j} + \vec{k})}{2} = \frac{2\vec{i} + 2\vec{k}}{2} = \vec{i} + \vec{k} \]

### Step 4: Calculate the Radius of the Sphere
The radius \( r \) of the sphere with diameter \( AB \) is half the length of \( AB \):
\[ |\vec{AB}| = |B - A| = |(-2\vec{j})| = 2 \]
Thus, the radius \( r \) is:
\[ r = \frac{|\vec{AB}|}{2} = \frac{2}{2} = 1 \]

### Step 5: Calculate the Distance from Point P to Midpoint M
Now, we need to find the distance from point \( P(2\vec{i} + 3\vec{j} + \vec{k}) \) to the midpoint \( M(\vec{i} + \vec{k}) \):
\[ \vec{PM} = P - M = (2\vec{i} + 3\vec{j} + \vec{k}) - (\vec{i} + \vec{k}) = (2\vec{i} - \vec{i}) + (3\vec{j} - 0) + (1 - 1)\vec{k} = \vec{i} + 3\vec{j} \]
The distance \( d \) is:
\[ d = |\vec{PM}| = |\vec{i} + 3\vec{j}| = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 6: Compare Distance with Radius
Since \( d = \sqrt{10} \) and \( r = 1 \):
\[ \sqrt{10} > 1 \]
This indicates that point \( P \) lies outside the sphere.

### Conclusion
Both statements are correct:
- Statement 1 is true because point \( P \) lies outside the sphere with diameter \( AB \).
- Statement 2 is also true as it provides a general condition for any points \( A \) and \( B \).