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Step-by-Step Solution
Step 1: Express the points A and B using parameters
• The first given line is
\frac{x - 7}{3} = \frac{y - 1}{-1} = \frac{z + 2}{1} .
Let the common parameter be \lambda .
• Hence, a general point on this line can be written as:
x = 3\lambda + 7, \quad y = -\lambda + 1, \quad z = \lambda - 2.
• Therefore, the point of intersection (labeled A) is:
A(3\lambda + 7, \; -\lambda + 1, \; \lambda - 2).
• The second given line is
\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1} .
Let the common parameter be \mu .
• Hence, a general point on this line can be written as:
x = 2\mu, \quad y = 3\mu + 7, \quad z = \mu.
• Therefore, the point of intersection (labeled B) is:
B(2\mu, \; 3\mu + 7, \; \mu).
Step 2: Use the direction ratios of the line intersecting A and B
• The line passing through A and B must have direction ratios proportional to (1, -4, 2) (given in the problem).
• The direction ratios of AB are given by the differences in the coordinates of B and A:
\bigl[\,2\mu - (3\lambda + 7), \; (3\mu + 7) - (-\lambda + 1), \; \mu - (\lambda - 2)\bigr].
• Simplifying inside the brackets gives:
(2\mu - 3\lambda - 7,\; 3\mu + \lambda + 6,\; \mu - \lambda + 2).
Step 3: Form the system of equations using proportionality
• Since the direction ratios of AB must be proportional to (1,\, -4,\, 2) , we set up:
\frac{2\mu - 3\lambda - 7}{1}
\,=\, \frac{3\mu + \lambda + 6}{-4}
\,=\, \frac{\mu - \lambda + 2}{2}.
• This leads to two independent equations (by equating the first ratio with the second, and the second with the third). Let us denote these equations as (i) and (ii):
(i)
\frac{2\mu - 3\lambda - 7}{1} = \frac{3\mu + \lambda + 6}{-4},
(ii)
\frac{3\mu + \lambda + 6}{-4} = \frac{\mu - \lambda + 2}{2}.
Step 4: Solve for \lambda and \mu
• From equation (i):
2\mu - 3\lambda - 7 = \frac{3\mu + \lambda + 6}{-4} \times 1.
• From equation (ii):
\frac{3\mu + \lambda + 6}{-4} = \frac{\mu - \lambda + 2}{2}.
• Rewriting these into simpler forms (as done in the provided solution) typically yields a system like:
5\lambda - 3\mu = -16, \quad \lambda - 5\mu = 10.
• Solving these simultaneously, we get:
\lambda = -5, \quad \mu = -3.
Step 5: Determine the coordinates of A and B
• Substitute \lambda = -5 into the coordinates of A:
A(3(-5) + 7,\; -(-5) + 1,\; (-5) - 2) = (-15 + 7,\; 5 + 1,\; -5 - 2).
A = (-8,\; 6,\; -7).
• Substitute \mu = -3 into the coordinates of B:
B(2(-3),\; 3(-3) + 7,\; -3) = (-6,\; -9 + 7,\; -3).
B = (-6,\; -2,\; -3).
Step 6: Find the square of the distance between A and B
• The vector \overrightarrow{AB} is:
B - A = [\,(-6) - (-8),\,(-2) - 6,\,(-3) - (-7)\,].
= (2,\, -8,\, 4).
• Therefore, the distance AB is:
AB = \sqrt{(2)^2 + (-8)^2 + (4)^2}
= \sqrt{4 + 64 + 16}
= \sqrt{84}.
• Hence,
(AB)^2 = 84.
Final Answer
The square of the distance between A and B is 84 .
