Step 1: Determine the Coordinates of Point P
The first line is given in symmetric form: $\frac{x-1}{2} = \frac{y+1}{-3} = z$. Let the common ratio be $t$. A general point $Q$ on this line can be represented parametrically as:
$x = 1 + 2t$
$y = -1 - 3t$
$z = t$
So, $Q = (1+2t, -1-3t, t)$.
The distance between $Q$ and the point $A(1, -1, 0)$ is given as $4\sqrt{14}$. Using the distance formula:
Distance$^2 = ((1+2t)-1)^2 + ((-1-3t)-(-1))^2 + (t-0)^2$
Distance$^2 = (2t)^2 + (-3t)^2 + t^2 = 4t^2 + 9t^2 + t^2 = 14t^2$
We are given that Distance $= 4\sqrt{14}$, so Distance$^2 = (4\sqrt{14})^2 = 16 \times 14 = 224$. Equating the two expressions for Distance$^2$:
$14t^2 = 224$
$t^2 = \frac{224}{14} = 16$
$t = \pm 4$
We need the point $P(\alpha, \beta, \gamma)$ that is nearer to the origin $(0, 0, 0)$. Let's find the coordinates for $t=4$ and $t=-4$ and check their distances from the origin:
- For $t=4$: $Q_1 = (1+2(4), -1-3(4), 4) = (9, -13, 4)$. Distance from origin $= \sqrt{9^2 + (-13)^2 + 4^2} = \sqrt{81 + 169 + 16} = \sqrt{266}$.
- For $t=-4$: $Q_2 = (1+2(-4), -1-3(-4), -4) = (-7, 11, -4)$. Distance from origin $= \sqrt{(-7)^2 + 11^2 + (-4)^2} = \sqrt{49 + 121 + 16} = \sqrt{186}$.
Since $\sqrt{186} < \sqrt{266}$, the point nearer to the origin corresponds to $t=-4$. Therefore, the coordinates of point $P$ are $(\alpha, \beta, \gamma) = (-7, 11, -4)$.
Step 2: Calculate the Shortest Distance Between the Two Lines
The first line ($L_1$) passes through point $P(\alpha, \beta, \gamma) = (-7, 11, -4)$ and has direction vector $\vec{d_1} = \langle 1, 2, 3 \rangle$. Let $\vec{a_1} = \langle -7, 11, -4 \rangle$.
The second line ($L_2$) is given by $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$. It passes through point $A_2(-5, 10, 3)$ and has direction vector $\vec{d_2} = \langle 2, 1, 1 \rangle$. Let $\vec{a_2} = \langle -5, 10, 3 \rangle$.
The shortest distance $D$ between two skew lines is given by the formula:
$D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{||\vec{d_1} \times \vec{d_2}||}$
First, calculate the vector connecting the points on the lines:
$\vec{a_2} - \vec{a_1} = \langle -5 - (-7), 10 - 11, 3 - (-4) \rangle = \langle 2, -1, 7 \rangle$
Next, calculate the cross product of the direction vectors:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \mathbf{i}(2 \times 1 - 3 \times 1) - \mathbf{j}(1 \times 1 - 3 \times 2) + \mathbf{k}(1 \times 1 - 2 \times 2)
= \mathbf{i}(2 - 3) - \mathbf{j}(1 - 6) + \mathbf{k}(1 - 4)
= -\mathbf{i} + 5\mathbf{j} - 3\mathbf{k} = \langle -1, 5, -3 \rangle$
Calculate the magnitude of the cross product:
$||\vec{d_1} \times \vec{d_2}|| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$
Now, calculate the dot product of $(\vec{a_2} - \vec{a_1})$ and $(\vec{d_1} \times \vec{d_2})$:
$(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = \langle 2, -1, 7 \rangle \cdot \langle -1, 5, -3 \rangle
= (2)(-1) + (-1)(5) + (7)(-3) = -2 - 5 - 21 = -28$
Finally, substitute these values into the shortest distance formula:
$D = \frac{|-28|}{\sqrt{35}} = \frac{28}{\sqrt{35}}$
Simplify the result:
$D = \frac{28}{\sqrt{35}} = \frac{28 \times \sqrt{35}}{\sqrt{35} \times \sqrt{35}} = \frac{28\sqrt{35}}{35} = \frac{4\sqrt{35}}{5}$
To match the format of the options, rewrite the expression:
$D = 4 \times \frac{\sqrt{35}}{5} = 4 \times \sqrt{\frac{35}{5^2}} = 4 \times \sqrt{\frac{35}{25}} = 4\sqrt{\frac{7}{5}}$
This matches Option D.
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