The equation of plane which passes through the point of intersection of lines
`vecr=hati+2hatj+3hatk+lamda(3hati+hatj+2hatk)` and `vecr=3hati+hatj+2hatk+mu(hati+2hatj+3hatk)` where `lamda,muinR` and has the greastest distance from the origin is :

Options

A. `vecr.(7hati+2hatj+4hatk)=54`
B. `vecr.(5hati+4hatj+3hatk)=57`
C. `vecr.(3hati+4hatj+5hatk)=49`
D. `vecr.(4hati+3hatj+5hatk)=50`

Explanation

To solve the problem, we need to find the equation of a plane that passes through the point of intersection of two lines given in vector form. The lines are:

1. \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + \hat{j} + 2\hat{k}) \)
2. \( \vec{r} = 3\hat{i} + \hat{j} + 2\hat{k} + \mu(\hat{i} + 2\hat{j} + 3\hat{k}) \)

We will follow these steps to find the required plane:

### Step 1: Find the direction vectors and points on the lines

From the first line, we can identify:
- Point \( A = (1, 2, 3) \)
- Direction vector \( \vec{d_1} = (3, 1, 2) \)

From the second line, we can identify:
- Point \( B = (3, 1, 2) \)
- Direction vector \( \vec{d_2} = (1, 2, 3) \)