If `A(3i-2j-k), B(2i+3j-4k), C(-i+j+2k)" and "D(4i+5j+lambdak)` are coplanar points, then `lambda=` ________

Answer

To find the value of \(\lambda\) for the coplanar points \(A(3i-2j-k)\), \(B(2i+3j-4k)\), \(C(-i+j+2k)\), and \(D(4i+5j+\lambda k)\), we can use the condition for coplanarity of vectors. The vectors \( \overrightarrow{AB}, \overrightarrow{AC}, \) and \( \overrightarrow{AD} \) must satisfy the scalar triple product condition, which states that the scalar triple product must be zero.

### Step 1: Find the vectors \(\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}\)

1. **Calculate \(\overrightarrow{AB}\)**:
\[ \overrightarrow{AB} = B - A = (2i + 3j - 4k) - (3i - 2j - k) = (2 - 3)i + (3 + 2)j + (-4 + 1)k = -i + 5j - 3k \]

2. **Calculate \(\overrightarrow{AC}\)**:
\[ \overrightarrow{AC} = C - A = (-i + j + 2k) - (3i - 2j - k) = (-1 - 3)i + (1 + 2)j + (2 + 1)k = -4i + 3j + 3k \]

3. **Calculate \(\overrightarrow{AD}\)**:
\[ \overrightarrow{AD} = D - A = (4i + 5j + \lambda k) - (3i - 2j - k) = (4 - 3)i + (5 + 2)j + (\lambda + 1)k = i + 7j + (\lambda + 1)k \]

### Step 2: Set up the scalar triple product

The scalar triple product is given by:
\[ \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = 0 \]

### Step 3: Calculate \(\overrightarrow{AC} \times \overrightarrow{AD}\)

Using the determinant form:
\[ \overrightarrow{AC} \times \overrightarrow{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 3 & 3 \\ 1 & 7 & \lambda + 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 3 \\ 7 & \lambda + 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -4 & 3 \\ 1 & \lambda + 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -4 & 3 \\ 1 & 7 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 3 & 3 \\ 7 & \lambda + 1 \end{vmatrix} = 3(\lambda + 1) - 3 \cdot 7 = 3\lambda + 3 - 21 = 3\lambda - 18\)
2. \(\begin{vmatrix} -4 & 3 \\ 1 & \lambda + 1 \end{vmatrix} = -4(\lambda + 1) - 3 \cdot 1 = -4\lambda - 4 - 3 = -4\lambda - 7\)
3. \(\begin{vmatrix} -4 & 3 \\ 1 & 7 \end{vmatrix} = -4 \cdot 7 - 3 \cdot 1 = -28 - 3 = -31\)

Putting it all together:
\[ \overrightarrow{AC} \times \overrightarrow{AD} = (3\lambda - 18)\hat{i} + (4\lambda + 7)\hat{j} - 31\hat{k} \]

### Step 4: Calculate \(\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})\)

Now we compute:
\[ \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = (-i + 5j - 3k) \cdot ((3\lambda - 18)\hat{i} + (4\lambda + 7)\hat{j} - 31\hat{k}) \]

Calculating the dot product:
\[ = -1(3\lambda - 18) + 5(4\lambda + 7) - 3(-31) \]
\[ = -3\lambda + 18 + 20\lambda + 35 + 93 \]
\[ = (20\lambda - 3\lambda) + (18 + 35 + 93) = 17\lambda + 146 \]

### Step 5: Set the equation to zero

Setting the scalar triple product to zero:
\[ 17\lambda + 146 = 0 \]
\[ 17\lambda = -146 \]
\[ \lambda = -\frac{146}{17} \]

### Final Answer
\[ \lambda = -\frac{146}{17} \]