Find the value of `lambda` for which the vectors `vec(a), vec(b), vec(c)` are coplanar, where
(i) `vec(a)=(2hat(i)-hat(j)+hat(k)), vec(b) = (hat(i)+2hat(j)+3hat(k) ) and vec(c)=(3 hat(i)+lambda hat(j) + 5 hat (k))`
(ii) `vec(a)lambda hat(i)-10 hat(j)-5k^(2), vec(b) =-7hat(i)-5hat(j) and vec(c)= hat(i)--4hat(j)-3hat(k)`
(iii) `vec(a)=hat(i)-hat(j)+hat(k), vec(b)= 2hat( i) + hat(j)-hat(k) and vec(c)= lambda hat(i) - hat(j) + lambda hat(k)`

Answer

To find the value of \( \lambda \) for which the vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, we need to set the scalar triple product of the vectors equal to zero. The scalar triple product can be represented as the determinant of a matrix formed by the components of the vectors.

### Part (i)
Given:
- \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \vec{c} = 3\hat{i} + \lambda \hat{j} + 5\hat{k} \)

We set up the determinant as follows:

\[ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & \lambda & 5 \end{vmatrix} = 0 \]

Now, we compute the determinant:

1. **Expanding the determinant along the first row:**
\[ = 2 \begin{vmatrix} 2 & 3 \\ \lambda & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 3 \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & \lambda \end{vmatrix} \]

2. **Calculating the smaller 2x2 determinants:**
\[ = 2(2 \cdot 5 - 3 \cdot \lambda) + (1 \cdot 5 - 3 \cdot 3) + (1 \cdot \lambda - 2 \cdot 3) \]
\[ = 2(10 - 3\lambda) + (5 - 9) + (\lambda - 6) \]
\[ = 20 - 6\lambda - 4 + \lambda - 6 \]
\[ = 20 - 10 - 6\lambda + \lambda \]
\[ = 10 - 5\lambda \]

3. **Setting the determinant equal to zero:**
\[ 10 - 5\lambda = 0 \]
\[ 5\lambda = 10 \]
\[ \lambda = 2 \]

### Part (ii)
Given:
- \( \vec{a} = \lambda \hat{i} - 10 \hat{j} - 5 \hat{k} \)
- \( \vec{b} = -7 \hat{i} - 5 \hat{j} \)
- \( \vec{c} = \hat{i} - 4 \hat{j} - 3 \hat{k} \)

Setting up the determinant:

\[ \begin{vmatrix} \lambda & -10 & -5 \\ -7 & -5 & 0 \\ 1 & -4 & -3 \end{vmatrix} = 0 \]

1. **Expanding the determinant:**
\[ = \lambda \begin{vmatrix} -5 & 0 \\ -4 & -3 \end{vmatrix} - (-10) \begin{vmatrix} -7 & 0 \\ 1 & -3 \end{vmatrix} - 5 \begin{vmatrix} -7 & -5 \\ 1 & -4 \end{vmatrix} \]

2. **Calculating the smaller 2x2 determinants:**
\[ = \lambda(-5 \cdot -3 - 0 \cdot -4) + 10(-7 \cdot -3 - 0 \cdot 1) - 5(-7 \cdot -4 - -5 \cdot 1) \]
\[ = \lambda(15) + 10(21) - 5(28 - 5) \]
\[ = 15\lambda + 210 - 5(23) \]
\[ = 15\lambda + 210 - 115 \]
\[ = 15\lambda + 95 \]

3. **Setting the determinant equal to zero:**
\[ 15\lambda + 95 = 0 \]
\[ 15\lambda = -95 \]
\[ \lambda = -\frac{19}{3} \]

### Part (iii)
Given:
- \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \)
- \( \vec{b} = 2\hat{i} + \hat{j} - \hat{k} \)
- \( \vec{c} = \lambda \hat{i} - \hat{j} + \lambda \hat{k} \)

Setting up the determinant:

\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} = 0 \]

1. **Expanding the determinant:**
\[ = 1 \begin{vmatrix} 1 & -1 \\ -1 & \lambda \end{vmatrix} - (-1) \begin{vmatrix} 2 & -1 \\ \lambda & \lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ \lambda & -1 \end{vmatrix} \]

2. **Calculating the smaller 2x2 determinants:**
\[ = 1(\lambda - 1) + (2\lambda + 1) + (-2 - \lambda) \]
\[ = \lambda - 1 + 2\lambda + 1 - 2 - \lambda \]
\[ = 2\lambda - 2 \]

3. **Setting the determinant equal to zero:**
\[ 2\lambda - 2 = 0 \]
\[ 2\lambda = 2 \]
\[ \lambda = 1 \]

### Summary of Results
- For part (i), \( \lambda = 2 \)
- For part (ii), \( \lambda = -\frac{19}{3} \)
- For part (iii), \( \lambda = 1 \)