Using vectors show that the point A (-2, 3, 5), B (7, 0, -1), C (-3, -2, -5) and D (3, 4, 7) are such that AB and CD intersect at the point P (1, 2, 3).

To show that the lines AB and CD intersect at the point P (1, 2, 3) using vectors, we proceed as follows:

---

1. **Vector representation of the lines AB and CD:**

- Line AB passes through \( A(-2, 3, 5) \) and \( B(7, 0, -1) \):
The vector equation of line AB is:
\[
\vec{r}_{AB} = \vec{A} + t \cdot \vec{AB}
\]
Here, \( \vec{A} = (-2, 3, 5) \) and \( \vec{AB} = \vec{B} - \vec{A} = (7 + 2, 0 - 3, -1 - 5) = (9, -3, -6) \).

Thus, \( \vec{r}_{AB} = (-2, 3, 5) + t(9, -3, -6) \).

Expanding:
\[
\vec{r}_{AB} = (-2 + 9t, 3 - 3t, 5 - 6t)
\]

- Line CD passes through \( C(-3, -2, -5) \) and \( D(3, 4, 7) \):
The vector equation of line CD is:
\[
\vec{r}_{CD} = \vec{C} + s \cdot \vec{CD}
\]
Here, \( \vec{C} = (-3, -2, -5) \) and \( \vec{CD} = \vec{D} - \vec{C} = (3 + 3, 4 + 2, 7 + 5) = (6, 6, 12) \).

Thus, \( \vec{r}_{CD} = (-3, -2, -5) + s(6, 6, 12) \).

Expanding:
\[
\vec{r}_{CD} = (-3 + 6s, -2 + 6s, -5 + 12s)
\]

---

2. **Condition for intersection:**

For the lines AB and CD to intersect at the point P (1, 2, 3), their parametric equations must satisfy:
\[
(-2 + 9t, 3 - 3t, 5 - 6t) = (-3 + 6s, -2 + 6s, -5 + 12s)
\]

Equating components:
\[
-2 + 9t = -3 + 6s \tag{1}
\]
\[
3 - 3t = -2 + 6s \tag{2}
\]
\[
5 - 6t = -5 + 12s \tag{3}
\]

---

3. **Solving the system of equations:**

From equation (1):
\[
9t - 6s = -1 \tag{4}
\]

From equation (2):
\[
-3t - 6s = -5 \tag{5}
\]

From equation (3):
\[
-6t - 12s = -10 \tag{6}
\]

Simplify equation (6):
\[
-3t - 6s = -5 \tag{7}
\]

Notice that equation (5) and (7) are identical, so only two unique equations exist.

Using equations (4) and (5):
- From equation (5):
\[
-3t - 6s = -5 \implies t = \frac{-5 - 6s}{-3} \implies t = \frac{5 + 6s}{3} \tag{8}
\]

Substitute \( t \) from equation (8) into equation (4):
\[
9\left(\frac{5 + 6s}{3}\right) - 6s = -1
\]
Simplify:
\[
15 + 18s - 6s = -1
\]
\[
15 + 12s = -1
\]
\[
12s = -16 \implies s = -\frac{4}{3}
\]

Substitute \( s = -\frac{4}{3} \) into equation (8):
\[
t = \frac{5 + 6\left(-\frac{4}{3}\right)}{3}
\]
\[
t = \frac{5 - 8}{3} = \frac{-3}{3} = -1
\]

---

4. **Verify the intersection point:**

- For line AB, substitute \( t = -1 \) into \( \vec{r}_{AB} \):
\[
\vec{r}_{AB} = (-2 + 9(-1), 3 - 3(-1), 5 - 6(-1))
\]
\[
\vec{r}_{AB} = (-2 - 9, 3 + 3, 5 + 6) = (1, 2, 3)
\]

- For line CD, substitute \( s = -\frac{4}{3} \) into \( \vec{r}_{CD} \):
\[
\vec{r}_{CD} = (-3 + 6\left(-\frac{4}{3}\right), -2 + 6\left(-\frac{4}{3}\right), -5 + 12\left(-\frac{4}{3}\right))
\]
\[
\vec{r}_{CD} = (-3 - 8, -2 - 8, -5 - 16) = (1, 2, 3)
\]

Since \( \vec{r}_{AB} = \vec{r}_{CD} = (1, 2, 3) \), the lines AB and CD intersect at \( P(1, 2, 3) \).

---

5. **Conclusion:**

The point \( P(1, 2, 3) \) lies on both lines AB and CD, confirming that the lines intersect at \( P \).