Question 306 of 521MediumMCQ
If the equation of plane passing through the mirror image of a point (2,3,1) with respect to line 2x+1=1y−3=−1z+2 and containing the line 3x−2=21−y=1z+1 is αx+βy+γz=24, then α+β+γ is equal to
Solutions · 1
Official solution
Let point M is (2\[\lambda\] - 1, \[\lambda\] + 3, - \[\lambda\] - 2)
D.R.'s of AM line are < 2\[\lambda\] - 1 - 2, \[\lambda\] + 3 - 3, - λ−2−1>
= < 2λ−3,λ,−\[\lambda\] -3 >
AM \[ \bot \] line L1
\[ \therefore \] \[2(2\lambda - 3) + 1(\lambda ) - 1( - \lambda - 3) = 0\]
\[6\lambda = 3,\lambda = {1 \over 2}\] \[ \therefore \] \[M \equiv \left( {0,{7 \over 2},{{ - 5} \over 2}} \right)\]
M is mid-point of A \& B
\[M = {{A + B} \over 2}\]
B = 2M - A
B \[ \equiv \] (-2, 4, -6)
Now we have to find equation of plane passing through B(-2, 4, -6) \& also containing the line
\[{{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}\] ....... (1)
\[{{x - 2} \over 3} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}\]
Point P on line is (2, 1, -1)
\[{\overrightarrow b _2}\] of line L2 is 3, -2, 1
\[\overrightarrow n ||({\overrightarrow b _2} \times \overrightarrow {PB} )\]
\[{\overrightarrow b _2} = 3\widehat i - 2\widehat j + \widehat k\]
\[\overrightarrow {PB} = - 4\widehat i + 3\widehat j - 5\widehat k\]
\[\overrightarrow n = 7\widehat i + 11\widehat j + \widehat k\]
\[ \therefore \] equation of plane is \[\overrightarrow r \,.\,\overrightarrow n = \overrightarrow a \,.\,\overrightarrow n \]
\[\overrightarrow r \,.\,(7\widehat i + 11\widehat j + \widehat k) = ( - 2\widehat i + 4\widehat j - 6\widehat k).(7\widehat i + 11\widehat j + \widehat k)\]
7x + 11y + z = -14 + 44 -6
7x + 11y + z = 24
\[ \therefore \] \[\alpha\] = 7
\[\beta\] = 11
\[\gamma\] = 1
\[ \therefore \] \[\alpha\] + \[\beta\] + \[\gamma\] = 19
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