Vectors `veca = hati+2hatj+3hatk, vec b = 2hati-hatj+hatk and vecc= 3hati+hatj+4hatk` are so placed that the end point of one vector is the starting point of the next vector. Then the vectors are

Options

A. not coplanar
B. coplanar but cannot form a triangle
C. coplanar and form a triangle
D. coplanar and can form a right angled triangle.

Explanation

To solve the problem, we need to determine the relationship between the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) given their coordinates. We will check if they are coplanar and if they can form a triangle.

### Step 1: Define the vectors
Given:
\[ \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \vec{B} = 2\hat{i} - \hat{j} + \hat{k} \]
\[ \vec{C} = 3\hat{i} + \hat{j} + 4\hat{k} \]

### Step 2: Check if they form a triangle
To check if the vectors can form a triangle, we need to verify if:
\[ \vec{C} = \vec{A} + \vec{B} \]

Calculating \(\vec{A} + \vec{B}\):
\[ \vec{A} + \vec{B} = (\hat{i} + 2\hat{j} + 3\hat{k}) + (2\hat{i} - \hat{j} + \hat{k}) \]
\[ = (1 + 2)\hat{i} + (2 - 1)\hat{j} + (3 + 1)\hat{k} \]
\[ = 3\hat{i} + 1\hat{j} + 4\hat{k} \]
\[ = \vec{C} \]

Since \(\vec{C} = \vec{A} + \vec{B}\), the vectors can form a triangle.

### Step 3: Check for coplanarity
To check if the vectors are coplanar, we can use the scalar triple product. The vectors are coplanar if:
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \]

First, we need to calculate \(\vec{B} \times \vec{C}\):
\[ \vec{B} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}, \quad \vec{C} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} \]
\[ \vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 1 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ 1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 3 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \]
\[ = \hat{i}((-1)(4) - (1)(1)) - \hat{j}((2)(4) - (1)(3)) + \hat{k}((2)(1) - (-1)(3)) \]
\[ = \hat{i}(-4 - 1) - \hat{j}(8 - 3) + \hat{k}(2 + 3) \]
\[ = -5\hat{i} - 5\hat{j} + 5\hat{k} \]

Now, we calculate \(\vec{A} \cdot (\vec{B} \times \vec{C})\):
\[ \vec{A} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \]
\[ \vec{B} \times \vec{C} = \begin{pmatrix} -5 \\ -5 \\ 5 \end{pmatrix} \]
\[ \vec{A} \cdot (\vec{B} \times \vec{C}) = (1)(-5) + (2)(-5) + (3)(5) \]
\[ = -5 - 10 + 15 = 0 \]

Since the scalar triple product is zero, the vectors are coplanar.

### Conclusion
The vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) are coplanar and can form a triangle.