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Step-by-Step Detailed Solution
Step 1: Express the line l1 in parametric form
The line l1 is given by
$ \frac{x+5}{3} = \frac{y+4}{1} = \frac{z-\alpha}{-2}. $
Let the common parameter be $ \lambda $. Hence, the coordinates of any point on l1 can be written as:
$ x = 3\lambda - 5,\quad y = \lambda - 4,\quad z = -2\lambda + \alpha. $
Step 2: Use coplanarity condition to find $ \alpha $
The line l2 is the intersection of the planes:
$3 x + 2 y + z - 2 = 0$
$x - 3 y + 2 z - 13 = 0$
A plane containing l2 can be written as:
$ (3x + 2y + z - 2) + \mu (x - 3y + 2z - 13) = 0, $
where $ \mu $ is a real constant. For l1 to lie in this plane, its parametric coordinates
$ (3\lambda - 5,\, \lambda - 4,\, -2\lambda + \alpha) $
must satisfy the above plane equation for all $ \lambda $.
On simplifying, one obtains the condition $ \alpha = 7. $
Step 3: Update the parametric form of l1
With $ \alpha = 7 $, the line l1 becomes:
$ x = 3\lambda - 5,\quad y = \lambda - 4,\quad z = -2\lambda + 7. $
Step 4: Form the vector from a general point on l1 to Q
We want the point $ P(a,b,c) $ on l1 that is closest to $ Q(-4,-3,2). $
This occurs when the vector
$ \overrightarrow{PQ} $
is perpendicular to the direction vector of l1.
• Direction vector of l1:
$ \langle 3,\, 1,\, -2 \rangle. $
• Coordinates of a general point P on l1:
$ (3\lambda - 5,\, \lambda - 4,\, -2\lambda + 7). $
Then
$
\overrightarrow{PQ}
=
\langle -4 - (3\lambda - 5),\; -3 - (\lambda - 4),\; 2 - (-2\lambda + 7)\rangle.
$
Simplifying:
$
\overrightarrow{PQ}
=
\langle -4 - 3\lambda + 5,\; -3 - \lambda + 4,\; 2 + 2\lambda - 7\rangle
=
\langle -3\lambda + 1,\; -\lambda + 1,\; 2\lambda - 5\rangle.
$
Step 5: Apply the perpendicularity condition
For $ \overrightarrow{PQ} $ to be perpendicular to the direction vector
$ \langle 3,\,1,\,-2\rangle, $
their dot product must be zero:
$
\langle -3\lambda + 1,\; -\lambda + 1,\; 2\lambda - 5\rangle
\cdot
\langle 3,\; 1,\; -2\rangle
=
0.
$
Expanding the dot product:
$
( -3\lambda + 1 ) \cdot 3
+
( -\lambda + 1 ) \cdot 1
+
(2\lambda - 5)\cdot(-2)
=
0.
$
Which becomes:
$
-9\lambda + 3
+
(-\lambda + 1)
+
(-4\lambda + 10)
=
0.
$
Combine like terms:
$
-9\lambda - \lambda - 4\lambda + 3 + 1 + 10 = 0
\implies
-14\lambda + 14 = 0
\implies
\lambda = 1.
$
Step 6: Compute $ P(a,b,c) $ and find $ |a| + |b| + |c| $
Substitute $ \lambda = 1 $ back into
$ (3\lambda - 5,\; \lambda - 4,\; -2\lambda + 7) $:
$ P = (3(1) - 5,\; 1 - 4,\; -2(1) + 7) = (-2,\; -3,\; 5). $
Finally,
$
|a| + |b| + |c|
=
|-2| + |-3| + |5|
=
2 + 3 + 5
=
10.
$
Therefore, the required sum is $10$.
