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

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$9$

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(B) The equation of the family of planes passing through the line of intersection of $x-2y-z-5=0$ and $x+y+3z-5=0$ is given by $(x-2y-z-5) + \lambda(x+y+3z-5) = 0$,which simplifies to $(1+\lambda)x + (-2+\lambda)y + (-1+3\lambda)z - (5+5\lambda) = 0$.The direction ratios of the line $x+y+2z-7=0=2x+3y+z-2$ are given by the cross product of the normals of the two planes:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(1-6) - \hat{j}(1-4) + \hat{k}(3-2) = -5\hat{i} + 3\hat{j} + \hat{k}$.Since the plane is parallel to this line,the normal to the plane must be perpendicular to the direction of the line:$(1+\lambda)(-5) + (-2+\lambda)(3) + (-1+3\lambda)(1) = 0$$-5 - 5\lambda - 6 + 3\lambda - 1 + 3\lambda = 0$$\lambda - 12 = 0 \Rightarrow \lambda = 12$.Substituting $\lambda = 12$ into the plane equation:$(1+12)x + (-2+12)y + (-1+36)z = 5(1+12)$$13x + 10y + 35z = 65$.Thus,$a=13, b=10, c=35$. The point is $(13, 10, 35)$.The distance of $(13, 10, 35)$ from the plane $2x+2y-z+16=0$ is:$d = \frac{|2(13) + 2(10) - 1(35) + 16|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|26 + 20 - 35 + 16|}{\sqrt{4+4+1}} = \frac{|27|}{3} = 9$.

Let the equation of the plane passing through the line $x-2y-z-5=0=x+y+3z-5$ and parallel to the line $x+y+2z-7=0=2x+3y+z-2$ be $ax+by+cz=65$. Then the distance of the point $(a, b, c)$ from the plane $2x+2y-z+16=0$ is $..........$.

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