To solve the problem, we need to find the point of intersection of the third line with the second line given the conditions of the problem. Let's break it down step by step.
### Step 1: Understand the given lines
1. The first line is given by the equations:
\[
x = y = z
\]
This can be represented as the parametric equations:
\[
(t, t, t) \quad \text{for } t \in \mathbb{R}
\]
2. The second line is given by:
\[
x = y/2 = z/3
\]
We can express this line in parametric form as:
\[
(s, 2s, 3s) \quad \text{for } s \in \mathbb{R}
\]
### Step 2: Identify the third line
The third line passes through the point (1, 1, 1). We can express this line in parametric form as:
\[
(1 + a, 1 + b, 1 + c)
\]
where \(a\), \(b\), and \(c\) are parameters that define the direction of the line.
### Step 3: Find the area of the triangle formed by the lines
The area of the triangle formed by the three lines can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times OA \times OB \times \sin(\theta)
\]
where \(O\) is the origin, \(A\) is a point on the first line, and \(B\) is a point on the second line.
### Step 4: Calculate the lengths OA and OB
1. For the first line, let \(t = 1\):
\[
A = (1, 1, 1)
\]
Thus, the length \(OA\) is:
\[
OA = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
\]
2. For the second line, let \(s = 1\):
\[
B = (1, 2, 3)
\]
Thus, the length \(OB\) is:
\[
OB = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}
\]
### Step 5: Find the angle between the lines
To find \(\sin(\theta)\), we need to calculate the cosine of the angle between the two lines. The direction vectors for the lines can be derived from their parametric forms.
1. Direction vector for the first line: \((1, 1, 1)\)
2. Direction vector for the second line: \((1, 2, 3)\)
Using the dot product:
\[
\cos(\theta) = \frac{(1, 1, 1) \cdot (1, 2, 3)}{|(1, 1, 1)| |(1, 2, 3)|}
\]
Calculating the dot product:
\[
(1 \cdot 1) + (1 \cdot 2) + (1 \cdot 3) = 1 + 2 + 3 = 6
\]
Calculating the magnitudes:
\[
|(1, 1, 1)| = \sqrt{3}, \quad |(1, 2, 3)| = \sqrt{14}
\]
Thus,
\[
\cos(\theta) = \frac{6}{\sqrt{3} \cdot \sqrt{14}} = \frac{6}{\sqrt{42}}
\]
Now, using \(\sin^2(\theta) + \cos^2(\theta) = 1\):
\[
\sin^2(\theta) = 1 - \left(\frac{6}{\sqrt{42}}\right)^2 = 1 - \frac{36}{42} = \frac{6}{42} = \frac{1}{7}
\]
So,
\[
\sin(\theta) = \frac{1}{\sqrt{7}}
\]
### Step 6: Calculate the area of the triangle
Now substituting into the area formula:
\[
\text{Area} = \frac{1}{2} \times \sqrt{3} \times \sqrt{14} \times \frac{1}{\sqrt{7}} = \frac{1}{2} \times \sqrt{42} \times \frac{1}{\sqrt{7}} = \frac{1}{2} \times \sqrt{6}
\]
Setting this equal to \(\sqrt{6}\):
\[
\frac{1}{2} \times \sqrt{6} = \sqrt{6}
\]
This gives us the value of \(s\) as \(2\).
### Step 7: Find the coordinates of the intersection point
Substituting \(s = 2\) into the parametric equations of the second line:
\[
B = (2, 4, 6)
\]
### Conclusion
The point of intersection of the third line with the second line is:
\[
\boxed{(2, 4, 6)}
\]
