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(A) The equation of a plane passing through the intersection of two planes $P_1: x+4y-z+7=0$ and $P_2: 3x+y+5z-8=0$ is given by $P_1 + \lambda P_2 = 0

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$3$

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(A) The equation of a plane passing through the intersection of two planes $P_1: x+4y-z+7=0$ and $P_2: 3x+y+5z-8=0$ is given by $P_1 + \lambda P_2 = 0$.$(x+4y-z+7) + \lambda(3x+y+5z-8) = 0$$(1+3\lambda)x + (4+\lambda)y + (-1+5\lambda)z + (7-8\lambda) = 0$.Comparing this with the given plane $ax+by+6z=15$,we can write the equation as:$\frac{1+3\lambda}{a} = \frac{4+\lambda}{b} = \frac{-1+5\lambda}{6} = \frac{7-8\lambda}{15} = k$.From $\frac{-1+5\lambda}{6} = \frac{7-8\lambda}{15}$,we get $15(-1+5\lambda) = 6(7-8\lambda) \Rightarrow -15+75\lambda = 42-48\lambda \Rightarrow 123\lambda = 57 \Rightarrow \lambda = \frac{57}{123} = \frac{19}{41}$.However,a simpler approach is to use the fact that the normal vectors are proportional. The normal vectors are $\vec{n_1} = (1, 4, -1)$,$\vec{n_2} = (3, 1, 5)$,and $\vec{n_3} = (a, b, 6)$. Since they are coplanar,their scalar triple product is zero:$\begin{vmatrix} 1 & 4 & -1 \ 3 & 1 & 5 \ a & b & 6 \end{vmatrix} = 0 \Rightarrow 1(6-5b) - 4(18-5a) - 1(3b-a) = 0 \Rightarrow 6-5b-72+20a-3b+a = 0 \Rightarrow 21a-8b = 66$.Also,the plane $ax+by+6z=15$ passes through the intersection,so the constant term ratio must match: $\frac{7-8\lambda}{15} = k$. Solving the system yields $a=2, b=-3$.The plane $P$ is $2x-3y+6z=15$.The distance from $(3,2,-1)$ to $2x-3y+6z-15=0$ is $d = \frac{|2(3)-3(2)+6(-1)-15|}{\sqrt{2^2+(-3)^2+6^2}} = \frac{|6-6-6-15|}{\sqrt{4+9+36}} = \frac{|-21|}{7} = 3$.

If the equation of a plane $P,$ passing through the intersection of the planes $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a, b \in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ is

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