The lines `(x-1)/(1)=(y-1)/(2)=(z-1)/(3)` and
`(x-4)/(2)=(y-6)/(3)=(z-7)/(3)` are coplanar. Their point of intersection is

Options

A. (`4,6,7)`
B. `(2,3,4)`
C. `(1,1,1)`
D. `(4,7,10)`

Answer

To find the point of intersection of the two lines given in the question, we will follow these steps:

### Step 1: Parameterize the Lines
The first line is given as:
\[ \frac{x-1}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(\lambda\):
\[ x = 1 + \lambda \]
\[ y = 1 + 2\lambda \]
\[ z = 1 + 3\lambda \]

The second line is given as:
\[ \frac{x-4}{2} = \frac{y-6}{3} = \frac{z-7}{3} = \mu \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(\mu\):
\[ x = 4 + 2\mu \]
\[ y = 6 + 3\mu \]
\[ z = 7 + 3\mu \]

### Step 2: Set the Equations Equal
Since both lines must intersect at the same point, we can set the equations for \(x\), \(y\), and \(z\) equal to each other.

From the \(x\) equations:
\[ 1 + \lambda = 4 + 2\mu \quad (1) \]

From the \(y\) equations:
\[ 1 + 2\lambda = 6 + 3\mu \quad (2) \]

From the \(z\) equations:
\[ 1 + 3\lambda = 7 + 3\mu \quad (3) \]

### Step 3: Solve the System of Equations
We will solve equations (1) and (2) first.

From equation (1):
\[ \lambda - 2\mu = 3 \quad (4) \]

From equation (2):
\[ 2\lambda - 3\mu = 5 \quad (5) \]

Now, we can solve equations (4) and (5) simultaneously.

### Step 4: Solve for \(\lambda\) and \(\mu\)
From equation (4):
\[ \lambda = 2\mu + 3 \]

Substituting this into equation (5):
\[ 2(2\mu + 3) - 3\mu = 5 \]
\[ 4\mu + 6 - 3\mu = 5 \]
\[ \mu + 6 = 5 \]
\[ \mu = -1 \]

Now substitute \(\mu = -1\) back into equation (4):
\[ \lambda = 2(-1) + 3 = 1 \]

### Step 5: Find the Point of Intersection
Now we can substitute \(\lambda = 1\) back into the equations for the first line:
\[ x = 1 + 1 = 2 \]
\[ y = 1 + 2(1) = 3 \]
\[ z = 1 + 3(1) = 4 \]

Thus, the point of intersection of the two lines is:
\[ (2, 3, 4) \]

### Final Answer
The point of intersection is:
\[ \boxed{(2, 3, 4)} \]