If ` vecp,vecq` and `vecr` are perpendicular to `vecq + vecr , vec r + vecp` and `vecp + vecq` respectively and if `|vecp +vecq| = 6, |vecq + vecr| = 4sqrt3` and `|vecr +vecp| = 4` then `|vecp + vecq + vecr|` is

Options

A. `5sqrt2`
B. 10
C. 15
D. 5

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations based on the perpendicular conditions
Given that:
1. \( \vec{p} \) is perpendicular to \( \vec{q} + \vec{r} \)
2. \( \vec{q} \) is perpendicular to \( \vec{p} + \vec{r} \)
3. \( \vec{r} \) is perpendicular to \( \vec{p} + \vec{q} \)

From these conditions, we can write the following equations:
- From \( \vec{p} \cdot (\vec{q} + \vec{r}) = 0 \):
\[ \vec{p} \cdot \vec{q} + \vec{p} \cdot \vec{r} = 0 \quad \text{(Equation 1)} \]

- From \( \vec{q} \cdot (\vec{p} + \vec{r}) = 0 \):
\[ \vec{q} \cdot \vec{p} + \vec{q} \cdot \vec{r} = 0 \quad \text{(Equation 2)} \]

- From \( \vec{r} \cdot (\vec{p} + \vec{q}) = 0 \):
\[ \vec{r} \cdot \vec{p} + \vec{r} \cdot \vec{q} = 0 \quad \text{(Equation 3)} \]

### Step 2: Use the given magnitudes
We are given:
- \( |\vec{p} + \vec{q}| = 6 \)
- \( |\vec{q} + \vec{r}| = 4\sqrt{3} \)
- \( |\vec{r} + \vec{p}| = 4 \)

We can express these magnitudes in terms of dot products:
1. \( |\vec{p} + \vec{q}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p} \cdot \vec{q} = 36 \) (Equation 4)
2. \( |\vec{q} + \vec{r}|^2 = |\vec{q}|^2 + |\vec{r}|^2 + 2\vec{q} \cdot \vec{r} = 48 \) (Equation 5)
3. \( |\vec{r} + \vec{p}|^2 = |\vec{r}|^2 + |\vec{p}|^2 + 2\vec{r} \cdot \vec{p} = 16 \) (Equation 6)

### Step 3: Solve the equations
Now we have a system of equations. Let's denote:
- \( a = |\vec{p}|^2 \)
- \( b = |\vec{q}|^2 \)
- \( c = |\vec{r}|^2 \)

From the equations we have:
1. \( a + b + 2\vec{p} \cdot \vec{q} = 36 \)
2. \( b + c + 2\vec{q} \cdot \vec{r} = 48 \)
3. \( c + a + 2\vec{r} \cdot \vec{p} = 16 \)

### Step 4: Substitute and simplify
Using Equations 1, 2, and 3, we can express \( \vec{p} \cdot \vec{q} \), \( \vec{q} \cdot \vec{r} \), and \( \vec{r} \cdot \vec{p} \) in terms of \( a \), \( b \), and \( c \):
- From Equation 1:
\[ \vec{p} \cdot \vec{q} = -\frac{a + b - 36}{2} \]
- From Equation 2:
\[ \vec{q} \cdot \vec{r} = -\frac{b + c - 48}{2} \]
- From Equation 3:
\[ \vec{r} \cdot \vec{p} = -\frac{c + a - 16}{2} \]

### Step 5: Add the equations
Adding the three equations gives:
\[ (a + b + c) + 2\left(-\frac{a + b - 36}{2} - \frac{b + c - 48}{2} - \frac{c + a - 16}{2}\right) = 36 + 48 + 16 \]

### Step 6: Solve for \( |\vec{p} + \vec{q} + \vec{r}| \)
Let \( x = |\vec{p} + \vec{q} + \vec{r}| \):
\[ x^2 = a + b + c + 2\left(\vec{p} \cdot \vec{q} + \vec{q} \cdot \vec{r} + \vec{r} \cdot \vec{p}\right) \]
Substituting the values we derived, we can find \( x^2 \) and thus \( x \).

### Final Answer
After calculations, we find:
\[ |\vec{p} + \vec{q} + \vec{r}| = 5\sqrt{2} \]