Find the image of the point `(0,2,3)` in the line `(x+3)/(5)= (y-1)/(2) = (z+4)/(3)`.

Answer

To find the image of the point \( P(0, 2, 3) \) in the line given by the equation

\[ \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}, \]

we will follow these steps:

### Step 1: Parametrize the line
The line can be expressed in parametric form. Let \( \lambda \) be the parameter. Then, we can write:

\[ x = 5\lambda - 3, \]
\[ y = 2\lambda + 1, \]
\[ z = 3\lambda - 4. \]

### Step 2: Find a point \( D \) on the line
Since \( D \) lies on the line, we can substitute \( \lambda \) into the equations to find the coordinates of point \( D \).

### Step 3: Find the direction ratios of line \( AD \)
Let \( A(0, 2, 3) \) be the point and \( D(x_D, y_D, z_D) \) be the point on the line. The direction ratios of line \( AD \) can be calculated as:

\[ \text{Direction ratios of } AD = (x_D - 0, y_D - 2, z_D - 3). \]

Substituting the expressions for \( x_D, y_D, z_D \):

\[ \text{Direction ratios of } AD = (5\lambda - 3, 2\lambda + 1 - 2, 3\lambda - 4 - 3) = (5\lambda - 3, 2\lambda - 1, 3\lambda - 7). \]

### Step 4: Direction ratios of the line
The direction ratios of the line given by the equation are \( (5, 2, 3) \).

### Step 5: Use the condition of perpendicularity
Since \( PD \) is perpendicular to the line \( L \), we can use the dot product condition:

\[ (5)(5\lambda - 3) + (2)(2\lambda - 1) + (3)(3\lambda - 7) = 0. \]

### Step 6: Solve for \( \lambda \)
Expanding the equation:

\[ 25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0, \]
\[ (25 + 4 + 9)\lambda - (15 + 2 + 21) = 0, \]
\[ 38\lambda - 38 = 0. \]

Thus, we find:

\[ \lambda = 1. \]

### Step 7: Substitute \( \lambda \) back to find \( D \)
Now substituting \( \lambda = 1 \) back into the parametric equations to find the coordinates of \( D \):

\[ x_D = 5(1) - 3 = 2, \]
\[ y_D = 2(1) + 1 = 3, \]
\[ z_D = 3(1) - 4 = -1. \]

So, \( D(2, 3, -1) \).

### Step 8: Use midpoint formula to find image \( Q \)
Since \( D \) is the midpoint of \( P \) and \( Q \), we can use the midpoint formula:

\[ D = \left( \frac{0 + \alpha}{2}, \frac{2 + \beta}{2}, \frac{3 + \gamma}{2} \right). \]

Setting the coordinates equal to those of \( D \):

1. \( \frac{0 + \alpha}{2} = 2 \) → \( \alpha = 4 \)
2. \( \frac{2 + \beta}{2} = 3 \) → \( \beta = 4 \)
3. \( \frac{3 + \gamma}{2} = -1 \) → \( \gamma = -5 \)

Thus, the image of the point \( P(0, 2, 3) \) is

\[ Q(4, 4, -5). \]

### Final Answer
The image of the point \( (0, 2, 3) \) in the line is \( (4, 4, -5) \).

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