To solve the problem of finding 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 follow these steps:
### Step 1: Identify the direction vectors and points on the lines
The first line can be expressed in parametric form as:
- \( x = 2 + 3t \)
- \( y = -4 + 2t \)
- \( z = 1 + 5t \)
From this, we can identify:
- Point on line 1: \( P_1(2, -4, 1) \)
- Direction vector of line 1: \( \vec{v_1} = (3, 2, 5) \)
The second line can be expressed in parametric form as:
- \( x = -1 - 2s \)
- \( y = 1 + 3s \)
- \( z = a + 4s \)
From this, we can identify:
- Point on line 2: \( P_2(-1, 1, a) \)
- Direction vector of line 2: \( \vec{v_2} = (-2, 3, 4) \)
### Step 2: Find the vector connecting the two points
The vector connecting the points \( P_1 \) and \( P_2 \) is given by:
\[
\vec{P_1P_2} = P_2 - P_1 = (-1 - 2, 1 + 4, a - 1) = (-3, 5, a - 1)
\]
### Step 3: Set up the determinant for coplanarity
The lines are coplanar if the scalar triple product of the vectors \( \vec{v_1} \), \( \vec{v_2} \), and \( \vec{P_1P_2} \) is zero. This can be expressed using the determinant:
\[
\begin{vmatrix}
3 & 2 & 5 \\
-2 & 3 & 4 \\
-3 & 5 & a - 1
\end{vmatrix} = 0
\]
### Step 4: Calculate the determinant
Calculating the determinant, we have:
\[
= 3 \begin{vmatrix}
3 & 4 \\
5 & a - 1
\end{vmatrix} - 2 \begin{vmatrix}
-2 & 4 \\
-3 & a - 1
\end{vmatrix} + 5 \begin{vmatrix}
-2 & 3 \\
-3 & 5
\end{vmatrix}
\]
Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 3 & 4 \\ 5 & a - 1 \end{vmatrix} = 3(a - 1) - 20 = 3a - 23 \)
2. \( \begin{vmatrix} -2 & 4 \\ -3 & a - 1 \end{vmatrix} = -2(a - 1) + 12 = -2a + 10 \)
3. \( \begin{vmatrix} -2 & 3 \\ -3 & 5 \end{vmatrix} = -10 + 9 = -1 \)
Substituting back into the determinant:
\[
3(3a - 23) - 2(-2a + 10) + 5(-1) = 0
\]
Expanding this gives:
\[
9a - 69 + 4a - 20 - 5 = 0
\]
Combining like terms:
\[
13a - 94 = 0
\]
### Step 5: Solve for \( a \)
Now, solving for \( a \):
\[
13a = 94 \implies a = \frac{94}{13}
\]
### Step 6: Find the greatest integer function \( [a] \)
Calculating \( \frac{94}{13} \):
\[
\frac{94}{13} \approx 7.230769
\]
Thus, the greatest integer function \( [a] = 7 \).
### Final Answer
\[
\boxed{7}
\]
