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1)

Let a, b and c be three real numbers satisfying [a b c]  [197827737]=[000]

If the point P(a, b, c), with reference to Eq. (E), lies on the plane 2x + y +z = 1, then the value of 7 a + b + c is


A) 0

B) 12

C) 7

D) 6

Answer:

Option D

Explanation:

 Given  ,[abc]1×3  [197827737]=[000]

  [a+8b+7c9a+2b+3c7a+7b+7c]=[000]

  a+8b+7c=0....(i)

  9a+2b+3c=0 .....(ii)

  a+b+c=0 .....(iii)

 On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get

 7a+c=0...(iv)

 Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get

 6a-b=0....(v)

   b=6a and c=-7a

 As  (a,b,c) lies on 2x+y+z=1

    2a+b+c=1

  2a+6a-7a=1

  a=1,b=6 and c=-7

    7a+b+c=7+6-7=6