Assertion : `If in a `/_\ABC, vec(BC)=vecp/|vecp|-vecq/|vecq| and vec(AC)=(2vecp)/|vecp|,|vecp|!=|veq|` then the value of `cos2A+cos2B+cos2C ` is -1., Reason: If in `/_\ABC, /_C=90^0 then cos2A+cos2B+cos2C=-1` (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

Options

A. Both statement I and statement II are correct and statement II is the correct explanation of statement I
B. both statement I and statement II are correct but statement II is not the correct explanation of statement I
C. Statement I is correct but statement II is incorrect
D. Statement II is correct but statement I is incorrect

Answer

To solve the given problem, we need to analyze both the assertion and the reason provided, and then determine their validity.

### Step 1: Analyze the Assertion
The assertion states that in triangle \( \triangle ABC \):
- \( \vec{BC} = \frac{\vec{p}}{|\vec{p}|} - \frac{\vec{q}}{|\vec{q}|} \)
- \( \vec{AC} = \frac{2\vec{p}}{|\vec{p}|} \)
- We need to prove that \( \cos 2A + \cos 2B + \cos 2C = -1 \).

### Step 2: Analyze the Reason
The reason states that if \( \angle C = 90^\circ \) in triangle \( \triangle ABC \), then \( \cos 2A + \cos 2B + \cos 2C = -1 \).

### Step 3: Prove the Reason
1. In triangle \( ABC \), since \( \angle C = 90^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \implies \angle A + \angle B = 90^\circ \]
2. We can express \( \cos 2A + \cos 2B + \cos 2C \):
\[ \cos 2C = \cos 180^\circ = -1 \]
3. Using the cosine addition formula:
\[ \cos 2A + \cos 2B = 2 \cos(A + B) \cos(A - B) \]
Since \( A + B = 90^\circ \):
\[ \cos(A + B) = \cos 90^\circ = 0 \]
Therefore:
\[ \cos 2A + \cos 2B = 0 \]
4. Thus:
\[ \cos 2A + \cos 2B + \cos 2C = 0 - 1 = -1 \]
This proves the reason is correct.

### Step 4: Prove the Assertion
Now we need to prove the assertion:
1. From the given vectors:
- \( \vec{BC} = \frac{\vec{p}}{|\vec{p}|} - \frac{\vec{q}}{|\vec{q}|} \)
- \( \vec{AC} = \frac{2\vec{p}}{|\vec{p}|} \)

2. Let:
- \( \vec{AB} = \vec{AC} + \vec{CB} \)
- \( \vec{AB} = \frac{2\vec{p}}{|\vec{p}|} + \left(-\frac{\vec{p}}{|\vec{p}|} + \frac{\vec{q}}{|\vec{q}|}\right) \)
- Simplifying gives:
\[ \vec{AB} = \frac{2\vec{p}}{|\vec{p}|} - \frac{\vec{p}}{|\vec{p}|} + \frac{\vec{q}}{|\vec{q}|} = \frac{\vec{p}}{|\vec{p}|} + \frac{\vec{q}}{|\vec{q}|} \]

3. Now, we can find \( \cos B \):
\[ \vec{BC} \cdot \vec{AB} = |\vec{BC}||\vec{AB}| \cos B \]
Using the magnitudes and properties of unit vectors, we can derive that \( \cos B = 0 \), leading to \( \angle B = 90^\circ \).

4. Thus, \( \angle A + \angle C = 90^\circ \) and we can conclude:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]
This proves the assertion is also correct.

### Final Conclusion
Both the assertion and the reason are true, but the reason is not the correct explanation for the assertion since we used different methods to prove them. Thus, the correct option is (B).