A line l passing through the origin is perpendicular to the lines
`1: (3 + t ) hati + (-1 +2 t ) hatj + (4 + 2t) hatk -oolt t lt ooand 1__(2) : (3 + 2s )hati + (3+2s) hati + (3+ 2s) hatj + ( 2+s) hatk , -oo lt s lt oo`
Then the coordinate(s) of the point(s) on `1 _(2)` at a distance of `sqrt17` from the point of intersection of 1 and `1_(1)` is (are)

Options

A. `((7)/(3), (7)/(3), (5)/(3))`
B. `(-1, -1, 0)`
C. `(1, 1, 1)`
D. `((7)/(9), (7)/(9), (8)/(9))`

Answer

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Identify the direction ratios of the lines L1 and L2

The equations of the lines are given as:
- Line L1: \( (3 + t) \hat{i} + (-1 + 2t) \hat{j} + (4 + 2t) \hat{k} \)
- Line L2: \( (3 + 2s) \hat{i} + (3 + 2s) \hat{j} + (2 + s) \hat{k} \)

From these equations, we can extract the direction ratios:
- For L1, the direction ratios are \( \hat{d_1} = (1, 2, 2) \)
- For L2, the direction ratios are \( \hat{d_2} = (2, 2, 1) \)

### Step 2: Find the direction ratios of line L

Since line L is perpendicular to both L1 and L2, we can find its direction ratios using the cross product of the direction ratios of L1 and L2.

\[ \hat{d_1} = (1, 2, 2), \quad \hat{d_2} = (2, 2, 1) \]

Calculating the cross product \( \hat{d_1} \times \hat{d_2} \):

\[ \hat{d_L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \hat{d_L} = \hat{i}(2 \cdot 1 - 2 \cdot 2) - \hat{j}(1 \cdot 1 - 2 \cdot 2) + \hat{k}(1 \cdot 2 - 2 \cdot 2) \]
\[ = \hat{i}(2 - 4) - \hat{j}(1 - 4) + \hat{k}(2 - 4) \]
\[ = -2\hat{i} + 3\hat{j} - 2\hat{k} \]

Thus, the direction ratios of line L are \( (-2, 3, -2) \).

### Step 3: Find the point of intersection of L and L1

Since line L passes through the origin, we can express it as:
\[ \text{Line L: } (0, 0, 0) + \lambda(-2, 3, -2) = (-2\lambda, 3\lambda, -2\lambda) \]

To find the intersection with line L1, we equate the coordinates:

\[ -2\lambda = 3 + t \quad (1) \]
\[ 3\lambda = -1 + 2t \quad (2) \]
\[ -2\lambda = 4 + 2t \quad (3) \]

From equation (1):
\[ t = -2\lambda - 3 \]

Substituting \( t \) into equation (2):
\[ 3\lambda = -1 + 2(-2\lambda - 3) \]
\[ 3\lambda = -1 - 4\lambda - 6 \]
\[ 3\lambda + 4\lambda = -7 \]
\[ 7\lambda = -7 \implies \lambda = -1 \]

Now substituting \( \lambda = -1 \) back into the equations to find \( t \):
\[ t = -2(-1) - 3 = 2 - 3 = -1 \]

Now substituting \( \lambda = -1 \) into the coordinates of L:
\[ M = (-2(-1), 3(-1), -2(-1)) = (2, -3, 2) \]

### Step 4: Find the coordinates on L2 at a distance of \( \sqrt{17} \) from point M

The distance formula between point M and a point on L2 \( (3 + 2s, 3 + 2s, 2 + s) \) is given by:

\[ \sqrt{(3 + 2s - 2)^2 + (3 + 2s + 3)^2 + (2 + s - 2)^2} = \sqrt{17} \]

Squaring both sides:
\[ (3 + 2s - 2)^2 + (3 + 2s + 3)^2 + (2 + s - 2)^2 = 17 \]

Simplifying:
\[ (1 + 2s)^2 + (6 + 2s)^2 + (s)^2 = 17 \]
\[ (1 + 4s + 4s^2) + (36 + 24s + 4s^2) + (s^2) = 17 \]
\[ 9s^2 + 28s + 20 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ s = \frac{-28 \pm \sqrt{28^2 - 4 \cdot 9 \cdot 20}}{2 \cdot 9} \]
\[ s = \frac{-28 \pm \sqrt{784 - 720}}{18} \]
\[ s = \frac{-28 \pm \sqrt{64}}{18} \]
\[ s = \frac{-28 \pm 8}{18} \]

Calculating the two possible values for \( s \):
1. \( s = \frac{-20}{18} = -\frac{10}{9} \)
2. \( s = \frac{-36}{18} = -2 \)

### Step 6: Find the coordinates for both values of \( s \)

For \( s = -2 \):
\[ P = (3 + 2(-2), 3 + 2(-2), 2 + (-2)) = (3 - 4, 3 - 4, 0) = (-1, -1, 0) \]

For \( s = -\frac{10}{9} \):
\[ P = \left(3 + 2\left(-\frac{10}{9}\right), 3 + 2\left(-\frac{10}{9}\right), 2 - \frac{10}{9}\right) \]
\[ = \left(3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}\right) \]
\[ = \left(\frac{27}{9} - \frac{20}{9}, \frac{27}{9} - \frac{20}{9}, \frac{18}{9} - \frac{10}{9}\right) \]
\[ = \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right) \]

### Final Answer

The coordinates of the points on L2 at a distance of \( \sqrt{17} \) from the point of intersection of L and L1 are:
1. \( (-1, -1, 0) \)
2. \( \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right) \)