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(B) The equation of the family of planes passing through the intersection of $P_1: 2x-y+z-3=0$ and $P_2: 4x-3y+5z+9=0$ is given by $P_1 + \lambda P_2

Vedclass · Accepted Answer

$-48$

Vedclass · Answer

(B) The equation of the family of planes passing through the intersection of $P_1: 2x-y+z-3=0$ and $P_2: 4x-3y+5z+9=0$ is given by $P_1 + \lambda P_2 = 0$. $(2x-y+z-3) + \lambda(4x-3y+5z+9) = 0$ $(2+4\lambda)x + (-1-3\lambda)y + (1+5\lambda)z + (-3+9\lambda) = 0$. Since the plane is parallel to the line with direction ratios $(2, 4, 5)$,the normal vector of the plane is perpendicular to the line's direction vector. Therefore,$2(2+4\lambda) + 4(-1-3\lambda) + 5(1+5\lambda) = 0$. $4 + 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$. $21\lambda + 5 = 0 \implies \lambda = -\frac{5}{21}$. Substituting $\lambda$ into the equation: $(2 - \frac{20}{21})x + (-1 + \frac{15}{21})y + (1 - \frac{25}{21})z + (-3 - \frac{45}{21}) = 0$. $(\frac{42-20}{21})x + (\frac{-21+15}{21})y + (\frac{21-25}{21})z + (\frac{-63-45}{21}) = 0$. $22x - 6y - 4z - 108 = 0$. Dividing by $2$: $11x - 3y - 2z - 54 = 0$. Here $\alpha=11, \beta=-3, \gamma=-2, d=-54$. $\alpha+\beta+\gamma+d = 11 - 3 - 2 - 54 = -48$.

The equation of the plane passing through the point of intersection of the planes $2x-y+z-3=0$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$. Then $\alpha+\beta+\gamma+d=$

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