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(C) Plane $\pi$ passes through $(3,-3,1)$ and is perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$.The direction ratios of the no

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$d$

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(C) Plane $\pi$ passes through $(3,-3,1)$ and is perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$.The direction ratios of the normal to the plane $\pi$ are $(3-2, 4-(-1), -1-5) = (1, 5, -6)$.Thus,the equation of plane $\pi$ is $1(x-3) + 5(y+3) - 6(z-1) = 0$,which simplifies to $x+5y-6z+18=0$.Let the second plane be $P_2: ax+y+cz-d=0$. This plane contains the points $(3,4,-1)$ and $(-1,2,5)$.Since $(3,4,-1)$ lies on $P_2$,we have $3a+4+c(-1)-d=0 \Rightarrow 3a-c-d=-4$ ... $(i)$.Since $(-1,2,5)$ lies on $P_2$,we have $-a+2+5c-d=0 \Rightarrow -a+5c-d=-2$ ... (ii).Subtracting (ii) from $(i)$,we get $4a-6c=-2 \Rightarrow 2a-3c=-1$ ... (iii).The normal vector of $P_2$ is $\vec{n_2} = (a, 1, c)$ and the normal vector of $\pi$ is $\vec{n_1} = (1, 5, -6)$.Since $P_2 \perp \pi$,their normals are perpendicular: $\vec{n_1} \cdot \vec{n_2} = 0 \Rightarrow a(1) + 1(5) + c(-6) = 0 \Rightarrow a-6c=-5$ ... (iv).From (iv),$a = 6c-5$. Substituting into (iii): $2(6c-5)-3c=-1 \Rightarrow 12c-10-3c=-1 \Rightarrow 9c=9 \Rightarrow c=1$.Then $a = 6(1)-5 = 1$.Substituting $a=1, c=1$ into $(i)$: $3(1)-1-d=-4 \Rightarrow 2-d=-4 \Rightarrow d=6$.Finally,$3(a+c) = 3(1+1) = 6$. Since $d=6$,$3(a+c) = d$.

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$

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