Find the image of the point `(3, -2, 1)` in the plane `3x-y+4x=2`

Answer

To find the image of the point \( (3, -2, 1) \) in the plane given by the equation \( 3x - y + 4z = 2 \), we can follow these steps:

### Step 1: Write the equation of the plane in standard form
The equation of the plane is given as:
\[ 3x - y + 4z - 2 = 0 \]
Here, we can identify \( a = 3 \), \( b = -1 \), \( c = 4 \), and \( d = -2 \).

### Step 2: Use the formula to find the image of the point
The formula to find the image of a point \( (x_1, y_1, z_1) \) with respect to the plane \( ax + by + cz + d = 0 \) is given by:
\[ \left( x, y, z \right) = \left( x_1, y_1, z_1 \right) + \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2} \left( a, b, c \right) \]

### Step 3: Substitute the values into the formula
Let \( (x_1, y_1, z_1) = (3, -2, 1) \).
Now, calculate \( ax_1 + by_1 + cz_1 + d \):
\[ = 3(3) + (-1)(-2) + 4(1) - 2 \]
\[ = 9 + 2 + 4 - 2 = 13 \]

Now, calculate \( a^2 + b^2 + c^2 \):
\[ = 3^2 + (-1)^2 + 4^2 = 9 + 1 + 16 = 26 \]

### Step 4: Calculate the image coordinates
Now substitute these values into the formula:
\[ \left( x, y, z \right) = (3, -2, 1) + \frac{-2(13)}{26} (3, -1, 4) \]
\[ = (3, -2, 1) + \frac{-26}{26} (3, -1, 4) \]
\[ = (3, -2, 1) + (-1)(3, -1, 4) \]
\[ = (3, -2, 1) + (-3, 1, -4) \]
\[ = (3 - 3, -2 + 1, 1 - 4) \]
\[ = (0, -1, -3) \]

### Conclusion
The image of the point \( (3, -2, 1) \) with respect to the plane \( 3x - y + 4z = 2 \) is:
\[ \boxed{(0, -1, -3)} \]