Find the co-ordinates of a point at which the line `(x+1)/(2) = (y-1)/(-2) = (z+5)/(6)`, meets the plane `x-2y+3z=8`.

Answer

To find the coordinates of the point at which the line \(\frac{x+1}{2} = \frac{y-1}{-2} = \frac{z+5}{6}\) meets the plane \(x - 2y + 3z = 8\), we can follow these steps:

### Step 1: Parametrize the Line
The given line can be expressed in parametric form. Let \(r\) be the parameter. Then we can write:
\[ x = 2r - 1, \quad y = -2r + 1, \quad z = 6r - 5 \]

### Step 2: Substitute into the Plane Equation
Since the point on the line must also satisfy the plane equation \(x - 2y + 3z = 8\), we substitute the parametric equations into the plane equation:
\[ (2r - 1) - 2(-2r + 1) + 3(6r - 5) = 8 \]

### Step 3: Simplify the Equation
Now, simplify the left-hand side:
\[ 2r - 1 + 4r - 2 + 18r - 15 = 8 \]
Combine like terms:
\[ (2r + 4r + 18r) + (-1 - 2 - 15) = 8 \]
This simplifies to:
\[ 24r - 18 = 8 \]

### Step 4: Solve for \(r\)
Now, solve for \(r\):
\[ 24r = 8 + 18 \]
\[ 24r = 26 \]
\[ r = \frac{26}{24} = \frac{13}{12} \]

### Step 5: Find the Coordinates
Now substitute \(r = \frac{13}{12}\) back into the parametric equations to find the coordinates:
1. For \(x\):
\[ x = 2\left(\frac{13}{12}\right) - 1 = \frac{26}{12} - \frac{12}{12} = \frac{14}{12} = \frac{7}{6} \]
2. For \(y\):
\[ y = -2\left(\frac{13}{12}\right) + 1 = -\frac{26}{12} + \frac{12}{12} = -\frac{14}{12} = -\frac{7}{6} \]
3. For \(z\):
\[ z = 6\left(\frac{13}{12}\right) - 5 = \frac{78}{12} - \frac{60}{12} = \frac{18}{12} = \frac{3}{2} \]

### Final Coordinates
Thus, the coordinates of the point where the line meets the plane are:
\[ \left(\frac{7}{6}, -\frac{7}{6}, \frac{3}{2}\right) \]