Statement 1: If `a` is an integer the the straight lines
`vecr=hati+2hati+3hatk+lamda(ahati+2hatj+3hatk)`
and `vecr=2hati+3hatj+hatk+mu(3hati+hatj+2hatk)` intersect at a point for `a=-5`.
Statement 2: Two straight lines intersect if the shortest distance between them is zero.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To determine whether the two given lines intersect at a point for \( a = -5 \), we can follow these steps:

### Step 1: Identify the lines
The two lines are given in vector form:
1. Line 1:
\[ \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(a\hat{i} + 2\hat{j} + 3\hat{k}) \]
Here, the point \( \vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k} \) and the direction vector \( \vec{b_1} = a\hat{i} + 2\hat{j} + 3\hat{k} \).

2. Line 2:
\[ \vec{r} = 2\hat{i} + 3\hat{j} + \hat{k} + \mu(3\hat{i} + a\hat{j} + 2\hat{k}) \]
Here, the point \( \vec{a_2} = 2\hat{i} + 3\hat{j} + \hat{k} \) and the direction vector \( \vec{b_2} = 3\hat{i} + a\hat{j} + 2\hat{k} \).

### Step 2: Find the vector connecting the points
Calculate \( \vec{a_2} - \vec{a_1} \):
\[ \vec{a_2} - \vec{a_1} = (2\hat{i} + 3\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k} \]

### Step 3: Calculate the cross product of the direction vectors
Now, calculate \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} = a\hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b_2} = 3\hat{i} + a\hat{j} + 2\hat{k} \]

Using the determinant to find the cross product:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a & 2 & 3 \\ 3 & a & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(2 \cdot 2 - 3 \cdot a) - \hat{j}(a \cdot 2 - 3 \cdot 3) + \hat{k}(a \cdot a - 2 \cdot 3) \]
\[ = \hat{i}(4 - 3a) - \hat{j}(2a - 9) + \hat{k}(a^2 - 6) \]

### Step 4: Find the shortest distance condition
The shortest distance between two lines is zero if:
\[ |\vec{a_2} - \vec{a_1}| \cdot (\vec{b_1} \times \vec{b_2}) = 0 \]

This means we need to set the dot product of \( \vec{a_2} - \vec{a_1} \) and \( \vec{b_1} \times \vec{b_2} \) to zero:
\[ (\hat{i} + \hat{j} - 2\hat{k}) \cdot ((4 - 3a)\hat{i} - (2a - 9)\hat{j} + (a^2 - 6)\hat{k}) = 0 \]

Calculating the dot product:
\[ 1(4 - 3a) + 1(-2a + 9) - 2(a^2 - 6) = 0 \]
\[ 4 - 3a - 2a + 9 - 2a^2 + 12 = 0 \]
\[ 25 - 5a - 2a^2 = 0 \quad \Rightarrow \quad 2a^2 + 5a - 25 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-25)}}{2 \cdot 2} \]
\[ = \frac{-5 \pm \sqrt{25 + 200}}{4} = \frac{-5 \pm 15}{4} \]
This gives:
\[ a = \frac{10}{4} = 2.5 \quad \text{or} \quad a = \frac{-20}{4} = -5 \]

### Conclusion
Thus, the lines intersect at \( a = -5 \) and \( a = 2.5 \). Therefore, Statement 1 is true, and Statement 2 is also true as it correctly describes the condition for intersection.