Let the plane P contain the line 2x+y-z-3=0=5 | vedclass

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(A) The equation of the family of planes passing through the line $2x+y-z-3=0$ and $5x-3y+4z+9=0$ is given by $(2x+y-z-3) + \lambda(5x-3y+4z+9) = 0$.T

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$26$

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(A) The equation of the family of planes passing through the line $2x+y-z-3=0$ and $5x-3y+4z+9=0$ is given by $(2x+y-z-3) + \lambda(5x-3y+4z+9) = 0$.This simplifies to $(2+5\lambda)x + (1-3\lambda)y + (-1+4\lambda)z + (9\lambda-3) = 0$.Since this plane is parallel to the line with direction vector $\vec{b} = (2, 4, 5)$,the normal vector of the plane $\vec{n} = (2+5\lambda, 1-3\lambda, -1+4\lambda)$ must be perpendicular to $\vec{b}$.Thus,$\vec{n} \cdot \vec{b} = 0 \implies 2(2+5\lambda) + 4(1-3\lambda) + 5(-1+4\lambda) = 0$.$4 + 10\lambda + 4 - 12\lambda - 5 + 20\lambda = 0 \implies 18\lambda + 3 = 0 \implies \lambda = -\frac{1}{6}$.Substituting $\lambda = -\frac{1}{6}$ into the plane equation: $(2 - \frac{5}{6})x + (1 + \frac{3}{6})y + (-1 - \frac{4}{6})z + (-\frac{9}{6} - 3) = 0$.Multiplying by $6$: $(12-5)x + (6+3)y + (-6-4)z + (-9-18) = 0 \implies 7x + 9y - 10z - 27 = 0$.The line passing through $A(8, -1, -19)$ parallel to $\frac{x}{-3} = \frac{y-5}{4} = \frac{2-z}{-12}$ (which is $\frac{x}{-3} = \frac{y-5}{4} = \frac{z-2}{12}$) has the equation $\frac{x-8}{-3} = \frac{y+1}{4} = \frac{z+19}{12} = k$.Any point on this line is $B(8-3k, -1+4k, -19+12k)$.For $B$ to lie on the plane $7x + 9y - 10z - 27 = 0$,we have $7(8-3k) + 9(-1+4k) - 10(-19+12k) - 27 = 0$.$56 - 21k - 9 + 36k + 190 - 120k - 27 = 0 \implies -105k + 210 = 0 \implies k = 2$.The distance $AB$ is the magnitude of the vector $\vec{AB} = (-3k, 4k, 12k)$ at $k=2$,which is $\sqrt{(-6)^2 + 8^2 + 24^2} = \sqrt{36 + 64 + 576} = \sqrt{676} = 26$.

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to $............$.

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