Let the lines L_{1}: overrightarrow{r} = lamb | vedclass

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(C) First,we find the intersection point $S$ of lines $L_{1}$ and $L_{2}$.For $L_{1}$,any point is $(\lambda, 2\lambda, 3\lambda)$.For $L_{2}$,any poi

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$5$

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(C) First,we find the intersection point $S$ of lines $L_{1}$ and $L_{2}$.For $L_{1}$,any point is $(\lambda, 2\lambda, 3\lambda)$.For $L_{2}$,any point is $(1+\mu, 3+\mu, 1+5\mu)$.Equating the coordinates: $\lambda = 1+\mu$,$2\lambda = 3+\mu$,$3\lambda = 1+5\mu$.From the first two: $\lambda - \mu = 1$ and $2\lambda - \mu = 3$. Subtracting gives $\lambda = 2$,so $\mu = 1$.Checking in the third: $3(2) = 6$ and $1+5(1) = 6$. Thus,$S = (2, 4, 6)$.The normal vector $\vec{n}$ to the plane is parallel to the cross product of the direction vectors of $L_{1}$ and $L_{2}$,which are $\vec{v}_{1} = \langle 1, 2, 3 angle$ and $\vec{v}_{2} = \langle 1, 1, 5 angle$.$\vec{n} = \vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & 1 & 5 \end{vmatrix} = \hat{i}(10-3) - \hat{j}(5-3) + \hat{k}(1-2) = 7\hat{i} - 2\hat{j} - \hat{k}$.The equation of the plane is $7(x-2) - 2(y-4) - 1(z-6) = 0$.$7x - 14 - 2y + 8 - z + 6 = 0 \Rightarrow 7x - 2y - z = 0$.Comparing with $ax + by - z + d = 0$,we get $a=7, b=-2, d=0$.Thus,$a + b + d = 7 - 2 + 0 = 5$.

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