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Three Dimensional Geometry

Question
CBSEENMA12033519

Show that the lines:
fraction numerator straight x minus 5 over denominator 4 end fraction space equals space fraction numerator straight y minus 7 over denominator 4 end fraction space equals fraction numerator straight z plus 3 over denominator negative 5 end fraction comma space space fraction numerator straight x minus 8 over denominator 7 end fraction space equals space fraction numerator straight y minus 4 over denominator 1 end fraction space equals space fraction numerator straight z minus 5 over denominator 3 end fraction are co-planar, find their common point and the equation of the plane in which way they lie. 

Solution
The equations of lines are
       fraction numerator straight x minus 5 over denominator 4 end fraction space equals space fraction numerator straight y minus 7 over denominator 4 end fraction space equals space fraction numerator straight z plus 3 over denominator negative 5 end fraction                      ...(1)
         fraction numerator straight x minus 8 over denominator 7 end fraction space equals fraction numerator straight y minus 4 over denominator 1 end fraction space equals space fraction numerator straight z minus 5 over denominator 3 end fraction                      ...(2)
Any point on the line (1) is (4 r + 5, 4 r + 7, – 5 r – 3)
It lies on the line (2), if fraction numerator 4 straight r plus 5 minus 8 over denominator 7 end fraction space equals space fraction numerator 4 straight r plus 7 minus 4 over denominator 1 end fraction space equals space fraction numerator negative 5 straight r minus 3 minus 5 over denominator 3 end fraction
i.e., if  fraction numerator 4 straight r minus 3 over denominator 7 end fraction equals fraction numerator 4 straight r plus 3 over denominator 1 end fraction space equals space fraction numerator negative 5 straight r minus 8 over denominator 3 end fraction                   ...(3)
Taking fraction numerator 4 straight r minus 3 over denominator 7 end fraction space equals fraction numerator 4 straight r plus 3 over denominator 1 end fraction comma we get
                         28 straight r plus 21 space equals space 4 straight r minus 3 comma space space space space space space space space space space space space space space space space space space space space therefore space space space space 24 straight r space equals space minus 24 comma space space space space space space space space space therefore space space space straight r space equals space minus 1
 Substituting this value of r in (3), we get,
            fraction numerator negative 4 minus 3 over denominator 7 end fraction space equals space fraction numerator negative 4 plus 3 over denominator 1 end fraction space equals space fraction numerator 5 minus 8 over denominator 3 end fraction

or – 1 = – 1 = – 1, which is true
∴  the lines intersect, and    ∴ are coplanar
The point of intersection of lines is (–4 + 5, –4 + 7, 5 – 3) i.e. (1,3,2)
The equation of plane in which given lines lie is
                           open vertical bar table row cell straight x minus 5 end cell cell straight y minus 7 end cell cell straight z plus 3 end cell row 4 4 cell negative 5 end cell row 7 1 3 end table close vertical bar space equals space 0
or     left parenthesis straight x minus 5 right parenthesis space open vertical bar table row 4 cell negative 5 end cell row 1 cell space space 3 end cell end table close vertical bar space minus space left parenthesis straight y minus 7 right parenthesis space open vertical bar table row 4 cell negative 5 end cell row 7 cell space space space 3 end cell end table close vertical bar plus space left parenthesis straight z plus 3 right parenthesis space open vertical bar table row 4 4 row 7 1 end table close vertical bar space equals space 0
or (x – 5) (12+ 5)–(y–7) (12 + 25) + (z + 3) (4 – 28) = 0
or    17 (x – 5) – 47 (y – 7) – 24 (z + 3) = 0
or    17x – 47 y – 24 z + 172 = 0
which is required equation of plane.