Show that the lines `(x+3)/(-3) = y - 1 = (z-5)/(5)` and `(x+1)/(-1) = (y-2)/(2) = (z-5)/(5)` are coplanar. Also find their point of intersection.

Answer

To show that the lines

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

and

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

are coplanar and to find their point of intersection, we can follow these steps:

### Step 1: Parametrize the lines

Let’s denote the first line by parameter \( \lambda \) and the second line by parameter \( k \).

For the first line, we can express the coordinates in terms of \( \lambda \):

\[ x_1 = -3\lambda - 3, \quad y_1 = \lambda + 1, \quad z_1 = 5\lambda + 5 \]

For the second line, we can express the coordinates in terms of \( k \):

\[ x_2 = -k - 1, \quad y_2 = 2k + 2, \quad z_2 = 5k + 5 \]

### Step 2: Set the coordinates equal to find intersection

To find the point of intersection, we set the coordinates equal:

\[ -3\lambda - 3 = -k - 1 \quad (1) \]
\[ \lambda + 1 = 2k + 2 \quad (2) \]
\[ 5\lambda + 5 = 5k + 5 \quad (3) \]

### Step 3: Solve the equations

From equation (3), we can simplify:

\[ 5\lambda + 5 = 5k + 5 \implies 5\lambda = 5k \implies \lambda = k \quad (4) \]

Now, substitute \( k \) from equation (4) into equations (1) and (2):

Substituting into equation (1):

\[ -3\lambda - 3 = -\lambda - 1 \]
\[ -3\lambda + \lambda = -1 + 3 \]
\[ -2\lambda = 2 \implies \lambda = -1 \]

Now substitute \( \lambda = -1 \) back into equation (4):

\[ k = -1 \]

### Step 4: Verify the solution

Now we have \( \lambda = -1 \) and \( k = -1 \). We can substitute these values back into the parametric equations to find the point of intersection.

Using \( \lambda = -1 \):

\[ x_1 = -3(-1) - 3 = 3 - 3 = 0 \]
\[ y_1 = -1 + 1 = 0 \]
\[ z_1 = 5(-1) + 5 = -5 + 5 = 0 \]

Using \( k = -1 \):

\[ x_2 = -(-1) - 1 = 1 - 1 = 0 \]
\[ y_2 = 2(-1) + 2 = -2 + 2 = 0 \]
\[ z_2 = 5(-1) + 5 = -5 + 5 = 0 \]

Both lines intersect at the point \( (0, 0, 0) \).

### Conclusion

The lines are coplanar since they intersect at the point \( (0, 0, 0) \).