A tetrahedron is three dimensional figure bounded by four non coplanar triangular plane. So a tetrahedron has points A,B,C,D as its vertices, which have coordinates `(x_(1),y_(1),z_(1)) (x_(2), y_(2), z_(2)) , (x _(3), y_(3) , z_(3)) and (x _(4), y _(4), z _(4))` respectively in a rectangular three –dimensional space. Then the coordinates of its centroid are
`((x_(1)+ x_(2) + x _(3) + x_(4))/(4) , (y _(1) + y _(2) + y_(3) + y _(4))/(4), (z_(1) + z_(2) + z_(3)+ z_(4))/(4)).`
The circumcentre of the tetrahedron is the centre of a sphere passing through its vertices. So, the circumcentre is a point equidistant from each of the vertices of tetrahedron.
Let tetrahedron has three of its vertices represented by the points
`(0,0,0) ,(6,-5,-1) and (-4,1,3)` and its centroid lies at the point `(1,-2,5).` Now answer the following questions
The equation of the triangular plane of tetrahedron that contains the given vertices is :

Options

A. `x-2y + z=0`
B. ` 5x - 3y - 2z =0`
C. `x+y+z=0`
D. `x+ 2y + 3z =0`

Answer

To find the equation of the triangular plane of the tetrahedron that contains the given vertices, we can follow these steps:

### Step 1: Identify the vertices of the tetrahedron
The vertices of the tetrahedron are given as:
- \( A(0, 0, 0) \)
- \( B(6, -5, -1) \)
- \( C(-4, 1, 3) \)

### Step 2: Write the general equation of the plane
The general equation of a plane in three-dimensional space can be expressed as:
\[ Ax + By + Cz + D = 0 \]
Since the plane passes through the point \( A(0, 0, 0) \), we can simplify this to:
\[ Ax + By + Cz = 0 \]

### Step 3: Substitute the coordinates of points B and C into the plane equation
We will substitute the coordinates of points \( B \) and \( C \) into the plane equation to form a system of equations.

1. For point \( B(6, -5, -1) \):
\[ 6A - 5B - C = 0 \quad \text{(Equation 1)} \]

2. For point \( C(-4, 1, 3) \):
\[ -4A + B + 3C = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
From Equation 1:
\[ C = 6A - 5B \]

Substituting \( C \) into Equation 2:
\[ -4A + B + 3(6A - 5B) = 0 \]
Expanding this gives:
\[ -4A + B + 18A - 15B = 0 \]
Combining like terms:
\[ (14A - 14B) = 0 \]
This simplifies to:
\[ A - B = 0 \quad \Rightarrow \quad A = B \]

### Step 5: Substitute \( A = B \) into the expression for \( C \)
Now substituting \( A = B \) into \( C = 6A - 5B \):
\[ C = 6A - 5A = A \]

### Step 6: Substitute \( A = B = C \) into the plane equation
Now substituting \( A = B = C \) into the plane equation \( Ax + By + Cz = 0 \):
\[ Ax + Ay + Az = 0 \quad \Rightarrow \quad A(x + y + z) = 0 \]
Since \( A \neq 0 \), we can divide by \( A \):
\[ x + y + z = 0 \]

### Final Answer
The equation of the triangular plane of the tetrahedron that contains the given vertices is:
\[ x + y + z = 0 \]

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