A line `l` passing through the origin is perpendicular to the lines `l_1: (3+t)hati+(-1+2t)hatj+(4+2t)hatk , -oo < t < oo , \ \ l_2: (3+s)hati+(3+2s)hatj+(2+s)hatk , -oo < t < oo` then the coordinates of the point on `l_2` at a distance of `sqrt17` from the point of intersection of `l&l_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))`

Explanation

To solve the problem step by step, we will follow the outlined process based on the given information about the lines \( l_1 \) and \( l_2 \), and the conditions provided.

### Step 1: Identify the direction vectors of lines \( l_1 \) and \( l_2 \)

The equations of the lines are given as:
- \( l_1: (3+t) \hat{i} + (-1+2t) \hat{j} + (4+2t) \hat{k} \)
- \( l_2: (3+s) \hat{i} + (3+2s) \hat{j} + (2+s) \hat{k} \)

From these equations, we can extract the direction vectors:
- For \( l_1 \), the direction vector \( \mathbf{d_1} = \hat{i} + 2\hat{j} + 2\hat{k} = (1, 2, 2) \)
- For \( l_2 \), the direction vector \( \mathbf{d_2} = \hat{i} + 2\hat{j} + \hat{k} = (1, 2, 1) \)

### Step 2: Find the direction vector of line \( l \)

Since line \( l \) is perpendicular to both \( l_1 \) and \( l_2 \), we can find its direction vector \( \mathbf{d} \) by taking the cross product of the direction vectors \( \mathbf{d_1} \) and \( \mathbf{d_2} \).

\[ \mathbf{d} = \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 1 & 2 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{d} = \hat{i} \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the minors:

\[ = \hat{i} (2 \cdot 1 - 2 \cdot 2) - \hat{j} (1 \cdot 1 - 1 \cdot 2) + \hat{k} (1 \cdot 2 - 1 \cdot 2) \]
\[ = \hat{i} (2 - 4) - \hat{j} (1 - 2) + \hat{k} (2 - 2) \]
\[ = -2\hat{i} + 1\hat{j} + 0\hat{k} \]

Thus, the direction vector of line \( l \) is \( (-2, 1, 0) \).

### Step 3: Find the point of intersection of lines \( l \) and \( l_1 \)

Let the point of intersection be \( M \) on line \( l_1 \). The parametric equation of line \( l \) through the origin can be expressed as:
\[ M = \lambda (-2, 1, 0) = (-2\lambda, \lambda, 0) \]

Setting this equal to the coordinates from line \( l_1 \):
\[ (-2\lambda, \lambda, 0) = (3+t, -1+2t, 4+2t) \]

This gives us the following equations:
1. \( -2\lambda = 3 + t \)
2. \( \lambda = -1 + 2t \)
3. \( 0 = 4 + 2t \)

From the third equation, we can solve for \( t \):
\[ 2t = -4 \implies t = -2 \]

Substituting \( t = -2 \) into the second equation:
\[ \lambda = -1 + 2(-2) = -1 - 4 = -5 \]

Now substituting \( \lambda = -5 \) into the first equation:
\[ -2(-5) = 10 \implies 10 = 3 + t \implies t = 7 \]

### Step 4: Find the coordinates of point \( M \)

Substituting \( t = -2 \) back into line \( l_1 \):
\[ M = (3 - 2, -1 + 2(-2), 4 + 2(-2)) = (1, -5, 0) \]

### Step 5: Find the coordinates of point \( P \) on line \( l_2 \)

We need to find point \( P \) on line \( l_2 \) such that the distance \( PM = \sqrt{17} \).

The coordinates of point \( P \) on line \( l_2 \) can be expressed as:
\[ P = (3+s, 3+2s, 2+s) \]

Using the distance formula:
\[ PM = \sqrt{(3+s - 1)^2 + (3+2s + 5)^2 + (2+s - 0)^2} = \sqrt{17} \]

Squaring both sides:
\[ (3+s - 1)^2 + (3+2s + 5)^2 + (2+s)^2 = 17 \]

This simplifies to:
\[ (2+s)^2 + (8+2s)^2 + (2+s)^2 = 17 \]

Expanding and combining like terms:
\[ (2+s)^2 + (8+2s)^2 + (2+s)^2 = 4 + 4s + s^2 + 64 + 32s + 4s^2 + 4 + 4s + s^2 = 17 \]
\[ 6s^2 + 40s + 68 = 17 \]
\[ 6s^2 + 40s + 51 = 0 \]

### Step 6: Solve the quadratic equation

Using the quadratic formula:
\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-40 \pm \sqrt{1600 - 1224}}{12} = \frac{-40 \pm \sqrt{376}}{12} \]
\[ = \frac{-40 \pm 2\sqrt{94}}{12} = \frac{-20 \pm \sqrt{94}}{6} \]

### Step 7: Find the coordinates of point \( P \)

Substituting the values of \( s \) back into the equation for \( P \):
1. For \( s_1 = \frac{-20 + \sqrt{94}}{6} \)
2. For \( s_2 = \frac{-20 - \sqrt{94}}{6} \)

Calculating the coordinates for each case will yield the points on line \( l_2 \) that are at a distance of \( \sqrt{17} \) from point \( M \).