Conisder the points (a-1, a, a + 1), (a, a + 1,a - 1) and (a + 1, a - 1, a).
1. These points always form the vertices of an equilateral triangle for any real value of a.
2. The area of the triangle formed by these points is independent of a.
Which of the statement (s) given above is/are correct ?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to analyze the given points and verify the two statements regarding the triangle formed by these points.

### Given Points:
1. \( A = (a-1, a, a+1) \)
2. \( B = (a, a+1, a-1) \)
3. \( C = (a+1, a-1, a) \)

### Step 1: Calculate the lengths of the sides of the triangle.

We will use the distance formula to find the lengths of the sides \( AB \), \( BC \), and \( AC \).

#### Length of \( AB \):
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of points \( A \) and \( B \):
\[ AB = \sqrt{(a - (a-1))^2 + ((a+1) - a)^2 + ((a-1) - (a+1))^2} \]
\[ = \sqrt{(1)^2 + (1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

#### Length of \( BC \):
\[ BC = \sqrt{((a+1) - a)^2 + ((a-1) - (a+1))^2 + (a - (a-1))^2} \]
\[ = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

#### Length of \( AC \):
\[ AC = \sqrt{((a+1) - (a-1))^2 + ((a-1) - a)^2 + ((a) - (a+1))^2} \]
\[ = \sqrt{(2)^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

### Conclusion for Step 1:
Since \( AB = BC = AC = \sqrt{6} \), the triangle formed by points \( A \), \( B \), and \( C \) is equilateral.

### Step 2: Calculate the area of the triangle.

The area \( A \) of a triangle formed by three points can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \]

#### Step 2.1: Find vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \vec{AB} = B - A = (a, a+1, a-1) - (a-1, a, a+1) = (1, 1, -2) \]
\[ \vec{AC} = C - A = (a+1, a-1, a) - (a-1, a, a+1) = (2, -1, -1) \]

#### Step 2.2: Calculate the cross product \( \vec{AB} \times \vec{AC} \):
Using the determinant:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & -1 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \left(1 \cdot (-1) - (-2) \cdot (-1)\right) - \hat{j} \left(1 \cdot (-1) - (-2) \cdot 2\right) + \hat{k} \left(1 \cdot (-1) - 1 \cdot 2\right) \]
\[ = \hat{i} (-1 - 2) - \hat{j} (-1 + 4) + \hat{k} (-1 - 2) \]
\[ = -3\hat{i} - 3\hat{j} - 3\hat{k} \]

#### Step 2.3: Calculate the magnitude of the cross product:
\[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{(-3)^2 + (-3)^2 + (-3)^2} = \sqrt{27} = 3\sqrt{3} \]

#### Step 2.4: Calculate the area:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{1}{2} \cdot 3\sqrt{3} = \frac{3\sqrt{3}}{2} \]

### Conclusion for Step 2:
The area of the triangle is \( \frac{3\sqrt{3}}{2} \), which is independent of \( a \).

### Final Answer:
Both statements are correct:
1. The points always form the vertices of an equilateral triangle for any real value of \( a \).
2. The area of the triangle formed by these points is independent of \( a \).