If ax+13y+bz+c=0 is a plane through the line of intersection of 2x+3y-z+1=0, x+y-2z+3=0 and is perpendicular to the plane 3x-y-2z=4, then the value of…

Question

If ax+13y+bz+c=0 is a plane through the line of intersection of 2x+3y-z+1=0, x+y-2z+3=0 and is perpendicular to the plane 3x-y-2z=4, then the value of 2a+3b+4c is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) -10. Equation of the plane through the line of intersection of given 2 planes is 2x+3y-z+1+λx+y-2z+3=0 2+λx+3+λy-1+2λz+1+3λ=0 ∵ it is perpendicular to 3x-y-2z=4 6+3λ-3-λ+4λ+2=0 6λ+5⇒λ=(-5)/6 7x+13y+4z-9=0 On comparing a=7,b=4,c=-9

If $a x + 13 y + b z + c = 0$ is a plane through the line of intersection of $2 x + 3 y - z + 1 = 0,$ $x + y - 2 z + 3 = 0$ and is perpendicular to the plane $3 x - y - 2 z = 4$ , then the value of $2 a + 3 b + 4 c$ is equal to

Solution

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