If equation of three lines are :
`(x)/(1)=(y)/(2)=(z)/(3), (x)/(2)=(y)/(1)=(z)/(3) and (x-1)/(1)=(2-y)/(1)=(z-3)/(0)`, then
which of the following statement(s) is/are correct ?

Options

A. Triangle formed by the line is equilateral
B. Triangle formed by the lines is isosceles
C. Equation of the plane containing the lines is `x+y=z`
D. Area of the triangle formed by the lines is `(3sqrt(3))/(2)`

Answer

To solve the problem, we will analyze the equations of the three lines given and find their intersection points. Then we will determine the properties of the triangle formed by these points.

### Step 1: Write the equations of the lines in parametric form

1. **First Line**:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} = \mu \]
From this, we can express the coordinates as:
\[ x = \mu, \quad y = 2\mu, \quad z = 3\mu \]

2. **Second Line**:
\[ \frac{x}{2} = \frac{y}{1} = \frac{z}{3} = \lambda \]
This gives us:
\[ x = 2\lambda, \quad y = \lambda, \quad z = 3\lambda \]

3. **Third Line**:
\[ \frac{x-1}{1} = \frac{2-y}{1} = \frac{z-3}{0} = \beta \]
This simplifies to:
\[ x = \beta + 1, \quad y = 2 - \beta, \quad z = 3 \]

### Step 2: Find the intersection points of the lines

1. **Intersection of the First and Second Lines**:
Set the coordinates equal:
\[ \mu = 2\lambda, \quad 2\mu = \lambda, \quad 3\mu = 3\lambda \]
From \(3\mu = 3\lambda\), we have \(\mu = \lambda\). Substituting \(\lambda\) in the first equation:
\[ \mu = 2\mu \implies \mu = 0 \implies \lambda = 0 \]
Therefore, the intersection point \(A\) is:
\[ A(0, 0, 0) \]

2. **Intersection of the Second and Third Lines**:
Set the coordinates equal:
\[ 2\lambda = \beta + 1, \quad \lambda = 2 - \beta, \quad 3\lambda = 3 \]
From \(3\lambda = 3\), we find \(\lambda = 1\). Substituting \(\lambda\) into the second equation:
\[ 1 = 2 - \beta \implies \beta = 1 \]
Thus, the intersection point \(B\) is:
\[ B(2, 1, 3) \]

3. **Intersection of the First and Third Lines**:
Set the coordinates equal:
\[ \mu = \beta + 1, \quad 2\mu = 2 - \beta, \quad 3\mu = 3 \]
From \(3\mu = 3\), we find \(\mu = 1\). Substituting \(\mu\) into the first equation:
\[ 1 = \beta + 1 \implies \beta = 0 \]
Thus, the intersection point \(C\) is:
\[ C(1, 2, 3) \]

### Step 3: Determine the properties of triangle ABC

We have the points:
- \(A(0, 0, 0)\)
- \(B(2, 1, 3)\)
- \(C(1, 2, 3)\)

1. **Calculate the distances**:
- Distance \(AB\):
\[ AB = \sqrt{(2-0)^2 + (1-0)^2 + (3-0)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \]
- Distance \(AC\):
\[ AC = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]
- Distance \(BC\):
\[ BC = \sqrt{(2-1)^2 + (1-2)^2 + (3-3)^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]

2. **Analyze the triangle**:
- Since \(AB = AC\) and \(BC\) is different, triangle \(ABC\) is an isosceles triangle.

### Step 4: Find the equation of the plane containing the triangle

The general equation of a plane is:
\[ Ax + By + Cz + D = 0 \]
We can use points \(A\), \(B\), and \(C\) to find the coefficients \(A\), \(B\), \(C\), and \(D\).

1. Substitute point \(A(0, 0, 0)\):
\[ D = 0 \]

2. Substitute point \(B(2, 1, 3)\):
\[ 2A + B + 3C = 0 \quad \text{(Equation 1)} \]

3. Substitute point \(C(1, 2, 3)\):
\[ A + 2B + 3C = 0 \quad \text{(Equation 2)} \]

4. Solve the equations:
Subtract Equation 1 from Equation 2:
\[ (A + 2B + 3C) - (2A + B + 3C) = 0 \implies -A + B = 0 \implies B = A \]
Substitute \(B = A\) in Equation 1:
\[ 2A + A + 3C = 0 \implies 3A + 3C = 0 \implies A + C = 0 \implies C = -A \]

Thus, we can express the plane as:
\[ x + y - z = 0 \quad \text{or} \quad x + y = z \]

### Step 5: Calculate the area of triangle ABC

Using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\theta) \]
We can find \(\sin(\theta)\) using the cosine rule:
\[ \cos(\theta) = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} \]
Substituting the distances:
\[ \cos(\theta) = \frac{14 + 14 - 2}{2 \cdot \sqrt{14} \cdot \sqrt{14}} = \frac{26}{28} = \frac{13}{14} \]
Then, using \(\sin^2(\theta) = 1 - \cos^2(\theta)\):
\[ \sin^2(\theta) = 1 - \left(\frac{13}{14}\right)^2 = 1 - \frac{169}{196} = \frac{27}{196} \implies \sin(\theta) = \frac{\sqrt{27}}{14} = \frac{3\sqrt{3}}{14} \]
Finally, substituting back to find the area:
\[ \text{Area} = \frac{1}{2} \cdot \sqrt{14} \cdot \sqrt{14} \cdot \frac{3\sqrt{3}}{14} = \frac{14 \cdot 3\sqrt{3}}{28} = \frac{3\sqrt{3}}{2} \]

### Conclusion

The statements regarding the triangle formed by the lines are:
1. The triangle is not equilateral.
2. The triangle is isosceles.
3. The equation of the plane containing the triangle is \(x + y = z\).
4. The area of the triangle is \(\frac{3\sqrt{3}}{2}\).