Let L be a line obtained from the intersection of two planes x+2y+z=6 and y+2z=4. If the point P(α,β,γ) is the foot of perpendicular from (3,2,1) on L, then the value of 21(α+β+γ) equals:
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Official solution
Dr's of line \[\left| {i10j21k12 } \right| = 3\widehat i - 2\widehat j + \widehat k\]
Dr/s : - (3, -2, 1)
Points on the line (-2, 4, 0)
Equation of the line \[{{x + 2} \over 3} = {{y - 4} \over { - 2}} = {z \over 1} = \lambda \]
Dr's of PQ : \[3\lambda - 5, - 2\lambda + 2,\lambda - 1\]
Dr's of y lines are (3, -2, 1)
Since \[PQ \bot \] line
\[3(3\lambda - 5) - 2( - 2\lambda + 2) + 1(\lambda - 1) = 0\]
\[\lambda = {{10} \over 7}\]
\[P\left( {{{16} \over 7},{8 \over 7},{{10} \over 7}} \right)\]
\[21(\alpha + \beta + \gamma ) = 21\left( {{{34} \over 7}} \right) = 102\]