`veca, vecb and vecc` are unit vecrtors such that `|veca + vecb+ 3vecc|=4` Angle between `veca and vecb is theta_(1)` , between `vecb and vecc is theta_(2)` and between `veca and vecb` varies `[pi//6, 2pi//3]` . Then the maximum value of `cos theta_(1)+3cos theta_(2)` is

Options

A. 3
B. 4
C. `2sqrt2`
D. 6

Explanation

To solve the problem step by step, we will follow the information given in the question and the video transcript.

### Step 1: Understand the given information
We have three unit vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
- \( |\vec{a} + \vec{b} + 3\vec{c}| = 4 \)
- The angles between the vectors are \( \theta_1 \) (between \( \vec{a} \) and \( \vec{b} \)) and \( \theta_2 \) (between \( \vec{b} \) and \( \vec{c} \)).
- The angle \( \theta_1 \) varies between \( \frac{\pi}{6} \) and \( \frac{2\pi}{3} \).

### Step 2: Square the magnitude equation
We start with the equation:
\[ |\vec{a} + \vec{b} + 3\vec{c}|^2 = 4^2 = 16 \]
Expanding the left-hand side:
\[ |\vec{a} + \vec{b} + 3\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |3\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(3\vec{c} \cdot \vec{a}) + 2(3\vec{c} \cdot \vec{b}) \]
Since \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \):
\[ = 1 + 1 + 9 + 2(\vec{a} \cdot \vec{b}) + 6(\vec{c} \cdot \vec{a}) + 6(\vec{c} \cdot \vec{b}) = 16 \]
This simplifies to:
\[ 11 + 2(\vec{a} \cdot \vec{b}) + 6(\vec{c} \cdot \vec{a}) + 6(\vec{c} \cdot \vec{b}) = 16 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b}) + 6(\vec{c} \cdot \vec{a}) + 6(\vec{c} \cdot \vec{b}) = 5 \]

### Step 4: Expressing in terms of cosines
Using the cosine of angles:
- \( \vec{a} \cdot \vec{b} = \cos(\theta_1) \)
- \( \vec{c} \cdot \vec{a} = \cos(\theta_3) \)
- \( \vec{c} \cdot \vec{b} = \cos(\theta_2) \)

Substituting these into the equation:
\[ 2\cos(\theta_1) + 6\cos(\theta_3) + 6\cos(\theta_2) = 5 \]

### Step 5: Finding maximum value of \( \cos(\theta_1) + 3\cos(\theta_2) \)
We want to maximize:
\[ \cos(\theta_1) + 3\cos(\theta_2) \]
From the rearranged equation:
\[ \cos(\theta_1) = \frac{5 - 6\cos(\theta_3) - 6\cos(\theta_2)}{2} \]
Substituting this into the expression we want to maximize:
\[ \frac{5 - 6\cos(\theta_3) - 6\cos(\theta_2)}{2} + 3\cos(\theta_2) \]
This simplifies to:
\[ \frac{5 - 6\cos(\theta_3)}{2} + \frac{6\cos(\theta_2)}{2} + 3\cos(\theta_2) = \frac{5 - 6\cos(\theta_3)}{2} + \frac{12\cos(\theta_2)}{2} \]
\[ = \frac{5 - 6\cos(\theta_3) + 12\cos(\theta_2)}{2} \]

### Step 6: Analyzing \( \cos(\theta_3) \)
To maximize this expression, we need to minimize \( \cos(\theta_3) \). The angle \( \theta_3 \) varies between \( \frac{\pi}{6} \) and \( \frac{2\pi}{3} \). The minimum value of \( \cos(\theta_3) \) occurs at \( \theta_3 = \frac{2\pi}{3} \):
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]

### Step 7: Substitute \( \cos(\theta_3) \) into the expression
Substituting \( \cos(\theta_3) = -\frac{1}{2} \):
\[ = \frac{5 - 6\left(-\frac{1}{2}\right) + 12\cos(\theta_2)}{2} = \frac{5 + 3 + 12\cos(\theta_2)}{2} = \frac{8 + 12\cos(\theta_2)}{2} = 4 + 6\cos(\theta_2) \]

### Step 8: Maximizing \( \cos(\theta_2) \)
The maximum value of \( \cos(\theta_2) \) is 1 (when \( \theta_2 = 0 \)):
\[ \text{Maximum value} = 4 + 6 \cdot 1 = 10 \]

### Final Answer
The maximum value of \( \cos(\theta_1) + 3\cos(\theta_2) \) is **10**.