Let the line frac{x-3}{7}=frac{y-2}{-1}=frac{ | vedclass

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(C) The plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$,which passes through $(4, -1, 0)$ and has direction vector $\vec{v_1} = \la

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$12$

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(C) The plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$,which passes through $(4, -1, 0)$ and has direction vector $\vec{v_1} = \langle 1, -2, 1 \rangle$. The plane also contains the line defined by $4ax-y+5z-7a=0$ and $2x-5y-z-3=0$. The family of planes passing through the intersection of these two planes is $(4ax-y+5z-7a) + \lambda(2x-5y-z-3) = 0$. Since the point $(4, -1, 0)$ lies on the plane,we have $(16a+1-7a) + \lambda(8+5-3) = 0$,which simplifies to $9a + 10\lambda + 1 = 0$ (Equation $1$). The normal vector of the plane is $\vec{n} = \langle 4a+2\lambda, -1-5\lambda, 5-\lambda \rangle$. Since the line $\vec{v_1}$ lies in the plane,$\vec{n} \cdot \vec{v_1} = 0$,so $(4a+2\lambda) - 2(-1-5\lambda) + (5-\lambda) = 0$,which simplifies to $4a + 11\lambda + 7 = 0$ (Equation $2$). Solving Equations $1$ and $2$ gives $a=1$ and $\lambda=-1$. Substituting these into the plane equation: $(4x-y+5z-7) - (2x-5y-z-3) = 0$,which simplifies to $2x+4y+6z-4=0$ or $x+2y+3z-2=0$. The line is $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}=t$,so any point on it is $(7t+3, -t+2, -4t+3)$. Substituting this into the plane equation: $(7t+3) + 2(-t+2) + 3(-4t+3) - 2 = 0$,which gives $7t+3-2t+4-12t+9-2=0$,so $-7t+14=0$,implying $t=2$. The point $P$ is $(17, 0, -5)$. Thus,$\alpha+\beta+\gamma = 17+0-5 = 12$.

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

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