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

Answer

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

### Step 1: Parametrize the line
We start by letting the common ratio be \(R\):
\[ \frac{x+1}{2} = \frac{y-3}{-2} = \frac{z+5}{6} = R \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(R\):
\[ x + 1 = 2R \implies x = 2R - 1 \]
\[ y - 3 = -2R \implies y = -2R + 3 \]
\[ z + 5 = 6R \implies z = 6R - 5 \]
Thus, the coordinates of the point \(Q\) on the line can be expressed as:
\[ Q(R) = (2R - 1, -2R + 3, 6R - 5) \]

### Step 2: Substitute into the plane equation
Next, we substitute these expressions for \(x\), \(y\), and \(z\) into the plane equation \(3x - y + z = 3\):
\[ 3(2R - 1) - (-2R + 3) + (6R - 5) = 3 \]

### Step 3: Simplify the equation
Now, we simplify the left-hand side:
\[ 6R - 3 + 2R - 3 + 6R - 5 = 3 \]
Combining like terms:
\[ (6R + 2R + 6R) - (3 + 3 + 5) = 3 \]
\[ 14R - 11 = 3 \]

### Step 4: Solve for \(R\)
Now, we solve for \(R\):
\[ 14R = 3 + 11 \]
\[ 14R = 14 \implies R = 1 \]

### Step 5: Find the coordinates of point \(Q\)
Now that we have \(R\), we can substitute it back into the expressions for \(x\), \(y\), and \(z\):
\[ x = 2(1) - 1 = 1 \]
\[ y = -2(1) + 3 = 1 \]
\[ z = 6(1) - 5 = 1 \]
Thus, the coordinates of point \(Q\) are:
\[ Q(1, 1, 1) \]

### Final Answer
The coordinates of the point at which the line meets the plane are:
\[ (1, 1, 1) \]