Step-by-Step Solution

1. Identify the given lines L1 and L2

The equations of the lines are given as:

Line L1: $r = \widehat i + \lambda ( - \widehat i + 2\widehat j + 2\widehat k)$ This line passes through the point $A(1, 0, 0)$ with position vector $\vec a_1 = \widehat i$. Its direction vector is $\vec b_1 = - \widehat i + 2\widehat j + 2\widehat k$.

Line L2: $r = \mu (2\widehat i - \widehat j + 2\widehat k)$ This line passes through the origin $O(0, 0, 0)$ with position vector $\vec a_2 = \vec 0$. Its direction vector is $\vec b_2 = 2\widehat i - \widehat j + 2\widehat k$.

2. Determine the direction vector of L3

The line L3 is perpendicular to both L1 and L2. Therefore, its direction vector, $\vec b_3$, must be parallel to the cross product of the direction vectors of L1 and L2, i.e., $\vec b_3 \parallel (\vec b_1 \times \vec b_2)$.

$\vec b_1 \times \vec b_2 = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix}$ $= \widehat i(2(2) - 2(-1)) - \widehat j((-1)(2) - 2(2)) + \widehat k((-1)(-1) - 2(2))$ $= \widehat i(4 + 2) - \widehat j(-2 - 4) + \widehat k(1 - 4)$ $= 6\widehat i + 6\widehat j - 3\widehat k$

We can take a simpler vector parallel to this one by dividing by 3. So, the direction vector for L3 can be taken as: $\vec b_3 = 2\widehat i + 2\widehat j - \widehat k$

All the given options have this direction vector, so we need to find a point on L3.

3. Find the points of intersection of L3 with L1 and L2

Since L3 cuts both L1 and L2, it must be the line of shortest distance between them. Let P be a point on L1 and Q be a point on L2 such that PQ is the shortest distance segment.

A general point P on L1 is given by the position vector: $\vec p = (\widehat i) + \lambda(-\widehat i + 2\widehat j + 2\widehat k) = (1-\lambda)\widehat i + (2\lambda)\widehat j + (2\lambda)\widehat k$

A general point Q on L2 is given by the position vector: $\vec q = \mu(2\widehat i - \widehat j + 2\widehat k) = (2\mu)\widehat i - (\mu)\widehat j + (2\mu)\widehat k$

The vector $\vec{PQ}$ is: $\vec{PQ} = \vec q - \vec p = (2\mu - (1-\lambda))\widehat i + (-\mu - 2\lambda)\widehat j + (2\mu - 2\lambda)\widehat k$

Since $\vec{PQ}$ is the shortest distance vector, it must be perpendicular to both $\vec b_1$ and $\vec b_2$.

Condition 1: $\vec{PQ} \cdot \vec b_1 = 0$ $(2\mu + \lambda - 1)(-1) + (-\mu - 2\lambda)(2) + (2\mu - 2\lambda)(2) = 0$ $-2\mu - \lambda + 1 - 2\mu - 4\lambda + 4\mu - 4\lambda = 0$ $(-2-2+4)\mu + (-1-4-4)\lambda + 1 = 0$ $-9\lambda + 1 = 0 \implies \lambda = \frac{1}{9}$

Condition 2: $\vec{PQ} \cdot \vec b_2 = 0$ $(2\mu + \lambda - 1)(2) + (-\mu - 2\lambda)(-1) + (2\mu - 2\lambda)(2) = 0$ $4\mu + 2\lambda - 2 + \mu + 2\lambda + 4\mu - 4\lambda = 0$ $(4+1+4)\mu + (2+2-4)\lambda - 2 = 0$ $9\mu - 2 = 0 \implies \mu = \frac{2}{9}$

Now we find the coordinates of points P and Q. Point P on L1 (for $\lambda = 1/9$): $\vec p = (1-\frac{1}{9})\widehat i + 2(\frac{1}{9})\widehat j + 2(\frac{1}{9})\widehat k = \frac{8}{9}\widehat i + \frac{2}{9}\widehat j + \frac{2}{9}\widehat k$

Point Q on L2 (for $\mu = 2/9$): $\vec q = 2(\frac{2}{9})\widehat i - (\frac{2}{9})\widehat j + 2(\frac{2}{9})\widehat k = \frac{4}{9}\widehat i - \frac{2}{9}\widehat j + \frac{4}{9}\widehat k$

4. Verify the options

Line L3 passes through points P and Q and has direction vector $\vec b_3 = 2\widehat i + 2\widehat j - \widehat k$. The equation of L3 can be written using any point on the line. We can write the equation of L3 as $r = \vec a + t \vec b_3$, where $\vec a$ is the position vector of any point on L3. Let's check the point given in each option.

Option A: $r = {2 \over 9}(2\widehat i - \widehat j + 2\widehat k) + t(2\widehat i + 2\widehat j - \widehat k)$ The point given is $\frac{2}{9}(2\widehat i - \widehat j + 2\widehat k) = \frac{4}{9}\widehat i - \frac{2}{9}\widehat j + \frac{4}{9}\widehat k$, which is the position vector of point Q. Since Q lies on L3, this equation represents L3. Option A is correct.

Option B: $r = {1 \over 3}(2\widehat i + k) + t(2\widehat i + 2\widehat j - \widehat k)$ The point given is $\vec a_B = \frac{2}{3}\widehat i + \frac{1}{3}\widehat k$. Let's check if this point lies on L3. The equation of L3 passing through Q is $r = (\frac{4}{9}\widehat i - \frac{2}{9}\widehat j + \frac{4}{9}\widehat k) + s(2\widehat i + 2\widehat j - \widehat k)$. Does $\vec a_B = \vec q + s \vec b_3$ for some scalar $s$? $\frac{2}{3}\widehat i + 0\widehat j + \frac{1}{3}\widehat k = (\frac{4}{9} + 2s)\widehat i + (-\frac{2}{9} + 2s)\widehat j + (\frac{4}{9} - s)\widehat k$ Comparing components:

• $\widehat i: \frac{2}{3} = \frac{4}{9} + 2s \implies \frac{6}{9} - \frac{4}{9} = 2s \implies \frac{2}{9} = 2s \implies s = \frac{1}{9}$
• $\widehat j: 0 = -\frac{2}{9} + 2s \implies \frac{2}{9} = 2s \implies s = \frac{1}{9}$
• $\widehat k: \frac{1}{3} = \frac{4}{9} - s \implies s = \frac{4}{9} - \frac{1}{3} = \frac{4-3}{9} = \frac{1}{9}$ Since we get a consistent value for $s$, the point lies on L3. Option B is correct.

Option C: $r = {2 \over 9}(4\widehat i + \widehat j + \widehat k) + t(2\widehat i + 2\widehat j - \widehat k)$ The point given is $\frac{2}{9}(4\widehat i + \widehat j + \widehat k) = \frac{8}{9}\widehat i + \frac{2}{9}\widehat j + \frac{2}{9}\widehat k$, which is the position vector of point P. Since P lies on L3, this equation represents L3. Option C is correct.

Option D: $r = t(2\widehat i + 2\widehat j - \widehat k)$ This line passes through the origin $(0,0,0)$. Let's check if the origin lies on L3. Does $(0,0,0) = (\frac{4}{9}, -\frac{2}{9}, \frac{4}{9}) + s(2, 2, -1)$?

• $0 = \frac{4}{9} + 2s \implies s = -\frac{2}{9}$
• $0 = -\frac{2}{9} + 2s \implies s = \frac{1}{9}$ We get a contradiction. The origin does not lie on L3. Option D is incorrect.