The image of the line `(x-1)/3=(y-3)/1=(z-4)/(-5)` in the plane `2x-y+z+3=0` is the line (1) `(x+3)/3=(y-5)/1=(z-2)/(-5)` (2) `(x+3)/(-3)=(y-5)/(-1)=(z+2)/5` (3) `(x-3)/3=(y+5)/1=(z-2)/(-5)` (3) `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`

Options

A. `(x+3)/(3)=(y-5)/(1)=(z-2)/(-5)`
B. `(x+3)/(-3)=(y-5)/(-1)=(z+2)/(5)`
C. `(x-3)/(3)=(y+5)/(1)=(z-2)/(-5)`
D. `(x-3)/(-3)=(y+5)/(-1)=(z-2)/(5)`

Answer

To find the image of the line given by the equation \((x-1)/3=(y-3)/1=(z-4)/(-5)\) in the plane \(2x - y + z + 3 = 0\), we will follow these steps:

### Step 1: Identify a point on the line
The line can be represented in parametric form:
\[ x = 1 + 3t, \quad y = 3 + t, \quad z = 4 - 5t \]
where \(t\) is a parameter.

### Step 2: Find a point on the plane
The equation of the plane is given as \(2x - y + z + 3 = 0\). We need to find a point on this plane. We can substitute \(x = 1\) and \(y = 3\) into the plane equation to find \(z\):
\[ 2(1) - 3 + z + 3 = 0 \implies 2 - 3 + z + 3 = 0 \implies z = -2 \]
Thus, the point \(P(1, 3, 4)\) lies on the line, and we can find the corresponding point \(M\) on the plane.

### Step 3: Find the coordinates of the foot of the perpendicular from the point to the plane
The coordinates of the foot of the perpendicular \(M\) from point \(P\) to the plane can be found using the formula for the projection of a point onto a plane. The normal vector of the plane \(2x - y + z + 3 = 0\) is \(\vec{n} = (2, -1, 1)\).

We can express point \(M\) as:
\[ M = P + k \vec{n} = (1, 3, 4) + k(2, -1, 1) \]
Substituting into the plane equation:
\[ 2(1 + 2k) - (3 - k) + (4 + k) + 3 = 0 \]
Simplifying this:
\[ 2 + 4k - 3 + k + 4 + k + 3 = 0 \implies 6k + 6 = 0 \implies k = -1 \]

### Step 4: Calculate the coordinates of point \(M\)
Substituting \(k = -1\) back into the equation for \(M\):
\[ M = (1, 3, 4) + (-1)(2, -1, 1) = (1 - 2, 3 + 1, 4 - 1) = (-1, 4, 3) \]

### Step 5: Find the image point \(Q\)
Let \(Q(x', y', z')\) be the image of point \(P\) across point \(M\). The midpoint \(M\) is given by:
\[ M = \left(\frac{1 + x'}{2}, \frac{3 + y'}{2}, \frac{4 + z'}{2}\right) \]
Setting this equal to the coordinates of \(M\):
\[ -1 = \frac{1 + x'}{2}, \quad 4 = \frac{3 + y'}{2}, \quad 3 = \frac{4 + z'}{2} \]

### Step 6: Solve for \(x', y', z'\)
1. For \(x'\):
\[ -1 = \frac{1 + x'}{2} \implies -2 = 1 + x' \implies x' = -3 \]
2. For \(y'\):
\[ 4 = \frac{3 + y'}{2} \implies 8 = 3 + y' \implies y' = 5 \]
3. For \(z'\):
\[ 3 = \frac{4 + z'}{2} \implies 6 = 4 + z' \implies z' = 2 \]

### Final Result
Thus, the coordinates of the image point \(Q\) are \((-3, 5, 2)\).

### Step 7: Write the equation of the image line
The image line can be expressed in symmetric form:
\[ \frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5} \]

### Conclusion
The correct option for the image of the line in the plane is:
\[ \frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5} \]