If the equation of the plane passing through the point (1 ,1 ,2 ) and perpendicular to the line x-3y+2z-1=04x-y+z os Ax+ By+ Cz=1, then 140 (C-B+A ) is equal to ________

Question

Accepted Answer

Correct answer: 15. (15) solution x-3y+2z-1=0 4x-y+z=0 n_1 n_2=(ijk1-324-11) =-i+7j+11k Dr’ of normal to the plane is -1 ,7 ,11 Equation of plane -1(x-1)+7(y-1)++11(z-2)=0 -x+7y+11z=28 -128x+ 7y28+ 11z28=1 Ax+ By+ Cz=1 140(C-B+A)=140( 1128- 728- 128) =140 328=15

$If the equation of the plane passing through the point (1, 1, 2) and perpendicular to the line x - 3 y + 2 z - 1 = 04 x - y + z os Ax + By + Cz = 1, then 140 (C - B + A) is equal to ________$

Solution

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