Question:
Let the equation of the plane, that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3 x-2 y+4 z-7=0$ and $x+5 y-2 z+9=0$, be $\alpha x+\beta y+\gamma z+3=0$, then $\alpha+\beta+\gamma$ is equal to :
Correct Option: 1
Solution:
Equation of plane is
$3 x-2 y+4 z-7+\lambda(x+5 y-2 z+9)=0$
$(3+\lambda) x+(5 \lambda-2) y+(4-2 \lambda) z+9 \lambda-7=0$
passing through $(1,4,-3)$
$\Rightarrow 3+\lambda+20 \lambda-8-12+6 \lambda+9 \lambda-7=0$
$\Rightarrow \lambda=\frac{2}{3}$
$\Rightarrow$ equation of plane is
$-11 x-4 y-8 z+3=0$
$\Rightarrow \alpha+\beta+\gamma=-23$
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