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Step-by-Step Solution
Step 1: Parametrize the given lines
ā¢ The first line, Lā, is given by
\frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}.
Introduce a parameter t_1 so that any point M on Lā can be written as:
M(\alpha, \beta, \gamma) = (1 + 3t_1,\; 2 + 2t_1,\; -1 - 2t_1).
Hence,
\alpha = 1 + 3t_1,\quad
\beta = 2 + 2t_1,\quad
\gamma = -1 - 2t_1.
ā¢ The second line, Lā, is given by
\frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}.
Introduce a parameter t_2 so that any point N on Lā can be written as:
N(a, b, c) = (-2 - 3t_2,\; 2 - 2t_2,\; 1 + 4t_2).
Hence,
a = -2 - 3t_2,\quad
b = 2 - 2t_2,\quad
c = 1 + 4t_2.
Step 2: Write down the direction vectors from point P to M and P to N
Let the external point be P(-1,\,2,\,3) .
ā¢ The vector from P to M is
(\alpha + 1,\; \beta - 2,\; \gamma - 3).
Substituting \alpha,\,\beta,\,\gamma :
( (1 + 3t_1) + 1,\; (2 + 2t_1) - 2,\; (-1 - 2t_1) - 3 )
= (2 + 3t_1,\; 2t_1,\; -4 - 2t_1).
ā¢ The vector from P to N is
(a + 1,\; b - 2,\; c - 3).
Substituting a,\,b,\,c :
( (-2 - 3t_2) + 1,\; (2 - 2t_2) - 2,\; (1 + 4t_2) - 3 )
= (-1 - 3t_2,\; -2t_2,\; -2 + 4t_2).
Step 3: Impose the condition that M and N lie on the same line through P
Because M and N lie on a line passing through P , their direction vectors (from P to M and P to N ) must be proportional:
(2 + 3t_1,\; 2t_1,\; -4 - 2t_1) \;\propto\; (-1 - 3t_2,\; -2t_2,\; -2 + 4t_2).
This gives the system of equations:
\frac{2 + 3t_1}{-1 - 3t_2}
= \frac{2t_1}{-2t_2}
= \frac{-4 - 2t_1}{-2 + 4t_2}.
Step 4: Solve for the parameters t_1 and t_2
From the ratios of the first two components:
\frac{2 + 3t_1}{-1 - 3t_2}
= \frac{2t_1}{-2t_2}.
By cross-multiplying the pairs of equations and simplifying (and similarly equating the second and third component ratios), we get:
t_1 = 4, \quad t_2 = 2.
Step 5: Find the coordinates of M and N
ā¢ Substituting t_1 = 4 into M(1 + 3t_1,\; 2 + 2t_1,\; -1 - 2t_1) :
M = (1 + 3 \cdot 4,\; 2 + 2 \cdot 4,\; -1 - 2 \cdot 4)
= (13,\; 10,\; -9).
Thus,
\alpha = 13,\quad \beta = 10,\quad \gamma = -9,\quad
\alpha + \beta + \gamma = 13 + 10 - 9 = 14.
ā¢ Substituting t_2 = 2 into N(-2 - 3t_2,\; 2 - 2t_2,\; 1 + 4t_2) :
N = (-2 - 3 \cdot 2,\; 2 - 2 \cdot 2,\; 1 + 4 \cdot 2)
= (-8,\; -2,\; 9).
Thus,
a = -8,\quad b = -2,\quad c = 9,\quad
a + b + c = -8 - 2 + 9 = -1.
Step 6: Compute the required ratio
We need the value of
\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2}.
Substituting the sums just calculated:
\frac{(14)^2}{(-1)^2} = \frac{196}{1} = 196.
Final Answer
The value of
\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}
is
196.
