The equation of plane which passes through the point of intersection of lines `vecr=hati+2hatj+3hatk+lambda(3hati+hatj+2hatk)` and `vecr=3hati+hatj+2hatk+mu(hati+2hatj+3hatk)` where `lambda, mu in R` and has the greatest 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`

Answer

To find the equation of the plane that passes through the point of intersection of the given lines and has the greatest distance from the origin, we will proceed step by step.

### Step 1: Write the equations of the lines in parametric form
The first line is given by:
\[ \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + \hat{j} + 2\hat{k}) \]
This can be rewritten as:
\[ \vec{r_1} = (1 + 3\lambda)\hat{i} + (2 + \lambda)\hat{j} + (3 + 2\lambda)\hat{k} \]

The second line is given by:
\[ \vec{r} = 3\hat{i} + \hat{j} + 2\hat{k} + \mu(\hat{i} + 2\hat{j} + 3\hat{k}) \]
This can be rewritten as:
\[ \vec{r_2} = (3 + \mu)\hat{i} + (1 + 2\mu)\hat{j} + (2 + 3\mu)\hat{k} \]