To find the value of \( [a] \) such that the lines
\[
\frac{x-2}{3} = \frac{y+4}{2} = \frac{z-1}{5}
\]
and
\[
\frac{x+1}{-2} = \frac{y-1}{3} = \frac{z-a}{4}
\]
are coplanar, we will use the condition that the scalar triple product of the direction ratios and the vector connecting points on the two lines must equal zero.
### Step 1: Identify the direction ratios of the lines
From the first line, we can extract the direction ratios:
- \( \vec{d_1} = (3, 2, 5) \)
From the second line, we extract the direction ratios:
- \( \vec{d_2} = (-2, 3, 4) \)
### Step 2: Identify points on the lines
From the first line, we can take a point:
- Point \( P_1 = (2, -4, 1) \)
From the second line, we can take a point:
- Point \( P_2 = (-1, 1, a) \)
### Step 3: Form the vector connecting the two points
The vector \( \vec{P_1P_2} \) from \( P_1 \) to \( P_2 \) is given by:
\[
\vec{P_1P_2} = P_2 - P_1 = (-1 - 2, 1 - (-4), a - 1) = (-3, 5, a - 1)
\]
### Step 4: Set up the scalar triple product
The scalar triple product of the vectors \( \vec{d_1}, \vec{d_2}, \vec{P_1P_2} \) must be zero for the lines to be coplanar:
\[
\vec{d_1} \cdot (\vec{d_2} \times \vec{P_1P_2}) = 0
\]
### Step 5: Calculate the cross product \( \vec{d_2} \times \vec{P_1P_2} \)
Using the determinant to calculate the cross product:
\[
\vec{d_2} \times \vec{P_1P_2} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 3 & 4 \\
-3 & 5 & a-1
\end{vmatrix}
\]
Calculating this determinant:
\[
= \hat{i} \left(3(a-1) - 20\right) - \hat{j} \left(-2(a-1) + 12\right) + \hat{k} \left(-2 \cdot 5 + 9\right)
\]
\[
= \hat{i} (3a - 3 - 20) - \hat{j} (-2a + 2 + 12) + \hat{k} (-10 + 9)
\]
\[
= \hat{i} (3a - 23) - \hat{j} (-2a + 14) + \hat{k} (-1)
\]
### Step 6: Calculate the scalar triple product
Now we compute \( \vec{d_1} \cdot (\vec{d_2} \times \vec{P_1P_2}) \):
\[
\vec{d_1} \cdot \left((3a - 23, 2a - 14, -1)\right) = 3(3a - 23) + 2(-2a + 14) + 5(-1)
\]
Expanding this gives:
\[
= 9a - 69 - 4a + 28 - 5 = 0
\]
Combining like terms:
\[
(9a - 4a) + (-69 + 28 - 5) = 0
\]
\[
5a - 46 = 0
\]
### Step 7: Solve for \( a \)
Solving for \( a \):
\[
5a = 46 \implies a = \frac{46}{5} = 9.2
\]
### Step 8: Find the greatest integer function value
Now we need to find \( [a] \):
\[
[a] = [9.2] = 9
\]
### Final Answer
Thus, the value of \( [a] \) is \( 9 \).
---
