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(C) The given equations of the lines are:$\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2} = \lambda$  .......$(1)$and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2

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$4x + 3y + 5z = 50$

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(C) The given equations of the lines are:$\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2} = \lambda$  .......$(1)$and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \mu$  ....$(2)$Any point on line $(1)$ is $P(3\lambda+1, \lambda+2, 2\lambda+3)$ and on line $(2)$ is $Q(\mu+3, 2\mu+1, 3\mu+2)$.For the point of intersection,we equate the coordinates:$3\lambda+1 = \mu+3 \implies 3\lambda - \mu = 2$$\lambda+2 = 2\mu+1 \implies \lambda - 2\mu = -1$Solving these equations,we get $\lambda=1$ and $\mu=1$.Substituting $\lambda=1$ in $P$,the point of intersection $R$ is $(4, 3, 5)$.The plane passing through $R(4, 3, 5)$ and having the largest distance from the origin $O(0, 0, 0)$ is the plane for which the vector $\vec{OR}$ is the normal vector.The normal vector $\vec{n} = \vec{OR} = 4\hat{i} + 3\hat{j} + 5\hat{k}$.The equation of the plane is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$,where $(a, b, c) = (4, 3, 5)$ and $(x_0, y_0, z_0) = (4, 3, 5)$.$4(x-4) + 3(y-3) + 5(z-5) = 0$$4x - 16 + 3y - 9 + 5z - 25 = 0$$4x + 3y + 5z = 50$.

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

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