Consider the line `L:vecr(hati+3hatj-hatk)+lamda(hatj+2hatk)` and the plane `pi:vecr(hati+4hatj+hatk)+6=0`
Statement 1: The line `L` intersects the plane `pi` at the point (1,0,-7).
Statement 2: The angle `theta` between the line `L` and the plane `pi` is given by `theta=1/2cos^(-1)(1/5)`.

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 solve the given problem, we need to analyze the two statements regarding the line \( L \) and the plane \( \pi \).

### Step 1: Find the equation of the line \( L \)
The line \( L \) is given by:
\[ \vec{r} = \hat{i} + 3\hat{j} - \hat{k} + \lambda(\hat{j} + 2\hat{k}) \]
This can be rewritten as:
\[ \vec{r} = \hat{i} + (3 + \lambda)\hat{j} + (-1 + 2\lambda)\hat{k} \]
Thus, a general point \( Q \) on the line can be expressed as:
\[ Q(1, 3 + \lambda, -1 + 2\lambda) \]

### Step 2: Write the equation of the plane \( \pi \)
The plane \( \pi \) is given by:
\[ \vec{r} \cdot (\hat{i} + 4\hat{j} + \hat{k}) + 6 = 0 \]
This can be expanded as:
\[ x + 4y + z + 6 = 0 \]

### Step 3: Substitute the coordinates of point \( Q \) into the plane equation
Substituting the coordinates of \( Q \) into the plane equation:
\[ 1 + 4(3 + \lambda) + (-1 + 2\lambda) + 6 = 0 \]
Simplifying this:
\[ 1 + 12 + 4\lambda - 1 + 2\lambda + 6 = 0 \]
\[ 18 + 6\lambda = 0 \]
Solving for \( \lambda \):
\[ 6\lambda = -18 \implies \lambda = -3 \]

### Step 4: Find the coordinates of the intersection point
Substituting \( \lambda = -3 \) back into the coordinates of \( Q \):
\[ Q(1, 3 - 3, -1 + 2(-3)) = Q(1, 0, -7) \]
Thus, the line \( L \) intersects the plane \( \pi \) at the point \( (1, 0, -7) \).

### Conclusion for Statement 1
Statement 1 is **true**.

### Step 5: Find the angle \( \theta \) between the line \( L \) and the plane \( \pi \)
To find the angle \( \theta \) between the line and the plane, we need the direction vector of the line and the normal vector of the plane.

- The direction vector \( \vec{b} \) of the line \( L \) is:
\[ \vec{b} = \hat{j} + 2\hat{k} \]
- The normal vector \( \vec{n} \) of the plane \( \pi \) is:
\[ \vec{n} = \hat{i} + 4\hat{j} + \hat{k} \]

### Step 6: Calculate the angle using the sine formula
The sine of the angle \( \theta \) between the line and the plane is given by:
\[ \sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} \]

Calculating \( \vec{b} \cdot \vec{n} \):
\[ \vec{b} \cdot \vec{n} = (0)(1) + (1)(4) + (2)(1) = 4 + 2 = 6 \]

Calculating the magnitudes:
\[ |\vec{b}| = \sqrt{0^2 + 1^2 + 2^2} = \sqrt{5} \]
\[ |\vec{n}| = \sqrt{1^2 + 4^2 + 1^2} = \sqrt{18} = 3\sqrt{2} \]

Substituting into the sine formula:
\[ \sin \theta = \frac{6}{\sqrt{5} \cdot 3\sqrt{2}} = \frac{6}{3\sqrt{10}} = \frac{2}{\sqrt{10}} = \frac{1}{5\sqrt{2}} \]

### Step 7: Relate sine to cosine
Using the identity \( \cos 2\theta = 1 - 2\sin^2 \theta \):
\[ \sin^2 \theta = \left(\frac{1}{5\sqrt{2}}\right)^2 = \frac{1}{50} \]
Thus,
\[ \cos 2\theta = 1 - 2\left(\frac{1}{50}\right) = 1 - \frac{2}{50} = \frac{48}{50} = \frac{24}{25} \]
This gives:
\[ \cos 2\theta = \frac{1}{5} \]
Thus,
\[ 2\theta = \cos^{-1}\left(\frac{1}{5}\right) \implies \theta = \frac{1}{2} \cos^{-1}\left(\frac{1}{5}\right) \]

### Conclusion for Statement 2
Statement 2 is also **true**.

### Final Answer
Both statements are true.