Co-ordiantes of the point of intersection of the line `(x+1)/1 = (y+3)/3 = (z-2)/-2` with the plane 3x+4y +5z=25 are

Options

A. (5,15,10)
B. (5,15,-10)
C. (5,-15,10)
D. (-5,-15,10)

Answer

To find the coordinates of the point of intersection of the line given by the equation

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

with the plane defined by

\[ 3x + 4y + 5z = 25, \]

we will follow these steps:

### Step 1: Parameterize the Line
Let

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

From this, we can express \(x\), \(y\), and \(z\) in terms of \(t\):

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

### Step 2: Substitute into the Plane Equation
Now we substitute \(x\), \(y\), and \(z\) into the plane equation \(3x + 4y + 5z = 25\):

\[ 3(t - 1) + 4(3t - 3) + 5(-2t + 2) = 25. \]

### Step 3: Simplify the Equation
Expanding this gives:

\[ 3t - 3 + 12t - 12 - 10t + 10 = 25. \]

Combining like terms:

\[ (3t + 12t - 10t) + (-3 - 12 + 10) = 25, \]
\[ 5t - 5 = 25. \]

### Step 4: Solve for \(t\)
Adding 5 to both sides:

\[ 5t = 30. \]

Dividing by 5:

\[ t = 6. \]

### Step 5: Find the Coordinates
Now substitute \(t = 6\) back into the expressions for \(x\), \(y\), and \(z\):

\[ x = 6 - 1 = 5, \]
\[ y = 3(6) - 3 = 18 - 3 = 15, \]
\[ z = -2(6) + 2 = -12 + 2 = -10. \]

### Conclusion
The coordinates of the point of intersection are

\[ (5, 15, -10). \]