The two lines `vecr=veca+veclamda(vecbxxvecc) and vecr=vecb+mu(veccxxveca)` intersect at a point where `veclamda and mu` are scalars then (A) `veca,vecb,vecc` are non coplanar (B) `|veca|=|vecb|=|vecc|` (C) `veca.vecc=vecb.vecc` (D) `lamda(vecbxvecc)+mu(veccxveca)=vecc`

Options

A. `vecaxxvecc=vecbxxvecc`
B. `veca.vecc=vecb.vecc`
C. `vecbxxveca=veccxxveca`
D. none of these

Explanation

To solve the problem, we start with the equations of the two lines given:

1. \(\vec{r} = \vec{a} + \lambda (\vec{b} \times \vec{c})\)
2. \(\vec{r} = \vec{b} + \mu (\vec{c} \times \vec{a})\)

Since the two lines intersect, we can set these two equations equal to each other:

\[ \vec{a} + \lambda (\vec{b} \times \vec{c}) = \vec{b} + \mu (\vec{c} \times \vec{a}) \]

### Step 1: Rearranging the Equation
Rearranging the equation gives us:

\[ \vec{a} - \vec{b} = \mu (\vec{c} \times \vec{a}) - \lambda (\vec{b} \times \vec{c}) \]

### Step 2: Define Cross Products
Let \(\vec{p} = \vec{b} \times \vec{c}\) and \(\vec{q} = \vec{c} \times \vec{a}\). The equation can be rewritten as:

\[ \vec{a} - \vec{b} = \mu \vec{q} - \lambda \vec{p} \]

### Step 3: Dot Product with \(\vec{c}\)
Now, we take the dot product of both sides with \(\vec{c}\):

\[ (\vec{a} - \vec{b}) \cdot \vec{c} = \mu (\vec{q} \cdot \vec{c}) - \lambda (\vec{p} \cdot \vec{c}) \]

### Step 4: Evaluating the Dot Products
Since \(\vec{p} = \vec{b} \times \vec{c}\) and \(\vec{q} = \vec{c} \times \vec{a}\), we know that:

- \(\vec{p} \cdot \vec{c} = 0\) (because \(\vec{p}\) is perpendicular to \(\vec{c}\))
- \(\vec{q} \cdot \vec{c} = 0\) (because \(\vec{q}\) is also perpendicular to \(\vec{c}\))

Thus, the equation simplifies to:

\[ (\vec{a} - \vec{b}) \cdot \vec{c} = 0 \]

### Step 5: Conclusion
This implies that:

\[ \vec{a} \cdot \vec{c} = \vec{b} \cdot \vec{c} \]

This means that the correct option is:

(C) \(\vec{a} \cdot \vec{c} = \vec{b} \cdot \vec{c}\)