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Step-by-Step Solution
Step 1: Express Each Given Line in Vector/Parametric Form
• The first line has the equations:
x = y + 2 = z.
Let the common parameter be \lambda . Then any point on this line can be written as:
P(\lambda) = (\lambda,\, \lambda - 2,\, \lambda).
• The second line has the equations:
x + 2 = 2y = 2z.
Let the common value be \mu . Then
2y = \mu \implies y = \frac{\mu}{2},
2z = \mu \implies z = \frac{\mu}{2},
and x + 2 = \mu \implies x = \mu - 2.
Hence, a point on this line can be written as:
Q(\mu) = (\mu - 2,\, \tfrac{\mu}{2},\, \tfrac{\mu}{2}).
Step 2: Use the Direction Ratios Condition
The line through P(\lambda) and Q(\mu) is given to have direction ratios proportional to (2,\,1,\,2).
Therefore,
Q(\mu) - P(\lambda) \;=\;
\Bigl[(\mu - 2) - \lambda,\;\tfrac{\mu}{2} - (\lambda - 2),\;\tfrac{\mu}{2} - \lambda\Bigr]
must be parallel to (2,\,1,\,2).
Hence, there exists a constant k such that:
(\mu - 2) - \lambda = 2k,\quad
\tfrac{\mu}{2} - (\lambda - 2) = k,\quad
\tfrac{\mu}{2} - \lambda = 2k.
Step 3: Solve for λ and μ
From the second equation:
\tfrac{\mu}{2} - (\lambda - 2) = k
\;\;\Longrightarrow\;\;
\tfrac{\mu}{2} - \lambda + 2 = k
\;\;\Longrightarrow\;\;
\tfrac{\mu}{2} - \lambda = k - 2.
But from the third equation we have:
\tfrac{\mu}{2} - \lambda = 2k.
Equate these two expressions for \tfrac{\mu}{2} - \lambda :
k - 2 = 2k \;\;\Longrightarrow\;\; k = -2.
Substitute k = -2 back into the third equation:
\tfrac{\mu}{2} - \lambda = 2k = -4.
Meanwhile, from the first equation:
(\mu - 2) - \lambda = 2k = -4
\;\;\Longrightarrow\;\; \mu - \lambda - 2 = -4
\;\;\Longrightarrow\;\; \mu - \lambda = -2.
We get the system:
\mu - \lambda = -2,
\quad
\tfrac{\mu}{2} - \lambda = -4.
From \mu - \lambda = -2 \implies \mu = \lambda - 2. Substituting into the second equation:
\tfrac{\lambda - 2}{2} - \lambda = -4
\;\;\Longrightarrow\;\;
\tfrac{\lambda}{2} - 1 - \lambda = -4
\;\;\Longrightarrow\;\;
-\tfrac{\lambda}{2} = -3
\;\;\Longrightarrow\;\;
\lambda = 6.
Therefore, \mu = 6 - 2 = 4.
Step 4: Determine Points P and Q
• Substituting \lambda = 6 into P(\lambda) :
P = (6,\,4,\,6).
• Substituting \mu = 4 into Q(\mu) :
Q = (2,\,2,\,2).
Step 5: Find the Vector PQ
\overrightarrow{PQ} = Q - P = (2-6,\;2-4,\;2-6)
= (-4,\;-2,\;-4).
Step 6: Compute the Perpendicular Distance from (1, 2, 12) to Line PQ
Let R = (1,\,2,\,12) . The formula for the perpendicular distance l from R to the line through P and Q is:
l =
\frac{\bigl|(\mathbf{R} - \mathbf{P}) \times \overrightarrow{PQ}\bigr|}{
\bigl|\overrightarrow{PQ}\bigr|
}.
• First, compute
\mathbf{R} - \mathbf{P} = (1-6,\;2-4,\;12-6) = (-5,\;-2,\;6).
• Next, find the cross product
(\mathbf{R} - \mathbf{P}) \times \overrightarrow{PQ}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-5 & -2 & 6 \\
-4 & -2 & -4
\end{vmatrix}.
Evaluating this determinant:
\[
= \mathbf{i}\bigl[(-2)(-4) - (6)(-2)\bigr]
\;-\; \mathbf{j}\bigl[(-5)(-4) - (6)(-4)\bigr]
\;+\; \mathbf{k}\bigl[(-5)(-2) - \bigl((-2)(-4)\bigr)\bigr].
\]
\[
= \mathbf{i}(8 + 12)
- \mathbf{j}(20 + 24)
+ \mathbf{k}(10 - 8)
\;=\;
20\mathbf{i} - 44\mathbf{j} + 2\mathbf{k}.
\]
Hence,
\bigl|(\mathbf{R} - \mathbf{P}) \times \overrightarrow{PQ}\bigr|
= \sqrt{20^2 + (-44)^2 + 2^2}
= \sqrt{400 + 1936 + 4}
= \sqrt{2340}
= 6\sqrt{65}.
Step 7: Find the Magnitude of PQ
\lvert \overrightarrow{PQ}\rvert
= \sqrt{(-4)^2 + (-2)^2 + (-4)^2}
= \sqrt{16 + 4 + 16}
= \sqrt{36}
= 6.
Step 8: Calculate the Required Distance and Then l^2
Using the distance formula:
l
= \frac{6\sqrt{65}}{6}
= \sqrt{65}.
Therefore,
l^2 = 65.
Hence, the value of l^2 is \boxed{65} .
