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(A) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.Here,$(3x-2y+4z

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$-23$

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(A) The equation of a plane passing through the intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.Here,$(3x-2y+4z-7) + \lambda(x+5y-2z+9) = 0$.Rearranging the terms,we get $(3+\lambda)x + (5\lambda-2)y + (4-2\lambda)z + (9\lambda-7) = 0$.Since this plane passes through the point $(1,4,-3)$,we substitute these coordinates into the equation:$(3+\lambda)(1) + (5\lambda-2)(4) + (4-2\lambda)(-3) + 9\lambda-7 = 0$.$3 + \lambda + 20\lambda - 8 - 12 + 6\lambda + 9\lambda - 7 = 0$.$36\lambda - 24 = 0 \Rightarrow 36\lambda = 24 \Rightarrow \lambda = \frac{24}{36} = \frac{2}{3}$.Substituting $\lambda = \frac{2}{3}$ back into the equation:$(3 + \frac{2}{3})x + (5(\frac{2}{3}) - 2)y + (4 - 2(\frac{2}{3}))z + (9(\frac{2}{3}) - 7) = 0$.$(\frac{11}{3})x + (\frac{4}{3})y + (\frac{8}{3})z - 1 = 0$.Multiplying by $-3$ to match the form $\alpha x + \beta y + \gamma z + 3 = 0$:$-11x - 4y - 8z + 3 = 0$.Comparing this with $\alpha x + \beta y + \gamma z + 3 = 0$,we get $\alpha = -11$,$\beta = -4$,and $\gamma = -8$.Therefore,$\alpha + \beta + \gamma = -11 - 4 - 8 = -23$.

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

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