Find the point where the line `(x-1)/2=(y-2)/-3=(z+3)/4` meets the plane `2x+4y-z=1`.

Answer

To find the point where the line given by the equation

\[ \frac{x-1}{2} = \frac{y-2}{-3} = \frac{z+3}{4} \]

meets the plane defined by

\[ 2x + 4y - z = 1, \]

we can follow these steps:

### Step 1: Parameterize the line
We can express the line in terms of a parameter \( \lambda \). From the equation of the line, we can write:

\[ x = 2\lambda + 1, \]
\[ y = -3\lambda + 2, \]
\[ z = 4\lambda - 3. \]

### Step 2: Substitute into the plane equation
Next, we substitute these expressions for \( x \), \( y \), and \( z \) into the plane equation:

\[ 2x + 4y - z = 1. \]

Substituting the parameterized values, we get:

\[ 2(2\lambda + 1) + 4(-3\lambda + 2) - (4\lambda - 3) = 1. \]

### Step 3: Simplify the equation
Now, we simplify the left-hand side:

\[ 4\lambda + 2 - 12\lambda + 8 - 4\lambda + 3 = 1. \]

Combining like terms:

\[ (4\lambda - 12\lambda - 4\lambda) + (2 + 8 + 3) = 1, \]
\[ -12\lambda + 13 = 1. \]

### Step 4: Solve for \( \lambda \)
Now, we solve for \( \lambda \):

\[ -12\lambda = 1 - 13, \]
\[ -12\lambda = -12, \]
\[ \lambda = 1. \]

### Step 5: Find the coordinates of the point
Now that we have \( \lambda = 1 \), we substitute this back into the parameterized equations for \( x \), \( y \), and \( z \):

\[ x = 2(1) + 1 = 3, \]
\[ y = -3(1) + 2 = -1, \]
\[ z = 4(1) - 3 = 1. \]

Thus, the point where the line meets the plane is

\[ (3, -1, 1). \]

### Final Answer
The point where the line meets the plane is

\[ \boxed{(3, -1, 1)}. \]