The point of intersection of the lines
`vec(r) = 7hat(i) + 10 hat(j)+ 13 hat(k)+vec(s)(2hat(i) + 3hat(j)+4hat(k))` and `vec(r) = 3hat(i) + 5hat(j) + 7hat(k) + t(hat(i) + 2hat(j) + 3hat(k))` is

Options

A. `hat(i) + hat(j) - hat(k)`
B. `2hat(i) - hat(j) + 4hat(k)`
C. `hat(i) - hat(j) + hat(k)`
D. `hat(i) + hat(j) + hat(k)`

Answer

To find the point of intersection of the given lines, we need to equate the two vector equations and solve for the parameters \( s \) and \( t \).

### Step 1: Write down the vector equations
The two lines are given as:
1. \( \vec{r_1} = 7\hat{i} + 10\hat{j} + 13\hat{k} + s(2\hat{i} + 3\hat{j} + 4\hat{k}) \)
2. \( \vec{r_2} = 3\hat{i} + 5\hat{j} + 7\hat{k} + t(\hat{i} + 2\hat{j} + 3\hat{k}) \)

### Step 2: Set the two equations equal to each other
To find the point of intersection, we set \( \vec{r_1} = \vec{r_2} \):
\[ 7\hat{i} + 10\hat{j} + 13\hat{k} + s(2\hat{i} + 3\hat{j} + 4\hat{k}) = 3\hat{i} + 5\hat{j} + 7\hat{k} + t(\hat{i} + 2\hat{j} + 3\hat{k}) \]

### Step 3: Expand both sides
Expanding both sides gives:
\[ (7 + 2s)\hat{i} + (10 + 3s)\hat{j} + (13 + 4s)\hat{k} = (3 + t)\hat{i} + (5 + 2t)\hat{j} + (7 + 3t)\hat{k} \]

### Step 4: Set up equations for each component
From the equality of coefficients, we can set up the following equations:
1. \( 7 + 2s = 3 + t \) (Equation 1)
2. \( 10 + 3s = 5 + 2t \) (Equation 2)
3. \( 13 + 4s = 7 + 3t \) (Equation 3)

### Step 5: Solve the equations simultaneously
From Equation 1:
\[ 2s - t = -4 \quad \Rightarrow \quad t = 2s + 4 \quad \text{(Equation 4)} \]

Substituting Equation 4 into Equation 2:
\[ 10 + 3s = 5 + 2(2s + 4) \]
Expanding gives:
\[ 10 + 3s = 5 + 4s + 8 \]
\[ 10 + 3s = 13 + 4s \]
Rearranging gives:
\[ 10 - 13 = 4s - 3s \quad \Rightarrow \quad -3 = s \]

Now substituting \( s = -3 \) back into Equation 4 to find \( t \):
\[ t = 2(-3) + 4 = -6 + 4 = -2 \]

### Step 6: Find the point of intersection
Now we can substitute \( s = -3 \) into the first line equation to find the point of intersection:
\[ \vec{r_1} = 7\hat{i} + 10\hat{j} + 13\hat{k} + (-3)(2\hat{i} + 3\hat{j} + 4\hat{k}) \]
Calculating this gives:
\[ = 7\hat{i} + 10\hat{j} + 13\hat{k} - 6\hat{i} - 9\hat{j} - 12\hat{k} \]
\[ = (7 - 6)\hat{i} + (10 - 9)\hat{j} + (13 - 12)\hat{k} \]
\[ = 1\hat{i} + 1\hat{j} + 1\hat{k} \]

Thus, the point of intersection is:
\[ \vec{P} = \hat{i} + \hat{j} + \hat{k} \]

### Final Answer
The point of intersection of the lines is \( \hat{i} + \hat{j} + \hat{k} \).