Show that the lines : `(x + 1)/(3) = (y + 3)/(5) = (z + 5)/(7)`
and `" " (x -2)/(1) = (y - 4)/(3) = (z -6)/(5)`
intersect each other. Also, find the their point of intersection.

Answer

To show that the lines intersect and to find the point of intersection, we will follow these steps:

### Step 1: Parameterize the Lines
The first line is given by:
\[ \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \]
Let this be equal to a parameter \(\lambda\). Then we can express \(x\), \(y\), and \(z\) in terms of \(\lambda\):
\[ x = 3\lambda - 1, \quad y = 5\lambda - 3, \quad z = 7\lambda - 5 \]

The second line is given by:
\[ \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} \]
Let this be equal to another parameter \(\mu\). Then we can express \(x\), \(y\), and \(z\) in terms of \(\mu\):
\[ x = \mu + 2, \quad y = 3\mu + 4, \quad z = 5\mu + 6 \]

### Step 2: Set the Equations Equal
To find the intersection, we need to equate the expressions for \(x\), \(y\), and \(z\):
1. From \(x\):
\[ 3\lambda - 1 = \mu + 2 \quad \text{(1)} \]
2. From \(y\):
\[ 5\lambda - 3 = 3\mu + 4 \quad \text{(2)} \]
3. From \(z\):
\[ 7\lambda - 5 = 5\mu + 6 \quad \text{(3)} \]

### Step 3: Solve the Equations
We will solve equations (1) and (2) first. Rearranging equation (1):
\[ 3\lambda - \mu = 3 \quad \text{(4)} \]

Rearranging equation (2):
\[ 5\lambda - 3\mu = 7 \quad \text{(5)} \]

Now we can solve equations (4) and (5) simultaneously. From equation (4), we can express \(\mu\) in terms of \(\lambda\):
\[ \mu = 3\lambda - 3 \]

Substituting this into equation (5):
\[ 5\lambda - 3(3\lambda - 3) = 7 \]
\[ 5\lambda - 9\lambda + 9 = 7 \]
\[ -4\lambda + 9 = 7 \]
\[ -4\lambda = -2 \quad \Rightarrow \quad \lambda = \frac{1}{2} \]

### Step 4: Find \(\mu\)
Now substituting \(\lambda = \frac{1}{2}\) back into equation (4) to find \(\mu\):
\[ \mu = 3\left(\frac{1}{2}\right) - 3 = \frac{3}{2} - 3 = -\frac{3}{2} \]

### Step 5: Find the Point of Intersection
Now we can find the point of intersection by substituting \(\lambda = \frac{1}{2}\) into the equations for \(x\), \(y\), and \(z\):
\[ x = 3\left(\frac{1}{2}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2} \]
\[ y = 5\left(\frac{1}{2}\right) - 3 = \frac{5}{2} - 3 = -\frac{1}{2} \]
\[ z = 7\left(\frac{1}{2}\right) - 5 = \frac{7}{2} - 5 = -\frac{3}{2} \]

Thus, the point of intersection is:
\[ \left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right) \]

### Conclusion
The lines intersect at the point \(\left(\frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}\right)\).