Plane $\pi$ passes through $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$.
$\therefore \mathrm{DR}$ 's of normal to the plane are
$(3-2),(4,+1),(-1-5)$
$\equiv(1,5,-6)$


Equation of plane $\pi$ is given by
$x+5 y-6 z+d=0$
$\because \pi$ consists the point $(3,-3,1)$.
$\begin{array}{ll}\Rightarrow & 3-15-6+d=0 \Rightarrow d=18 \\ \therefore & \pi \equiv x+5 y-6 z+18=0\end{array}$
Equation of plane containing points $(3,4,-1)$ and
$(-1,2,5)$ is $a x+y+c z-d=0$...(i)

Normal to this plane will be perpendicular to the line joining the points $(3,4,-1)$ and $(-1,2,5)$.
Then, $a(3+1)+1(4-2)+c(-1-5)=0$
$\Rightarrow \quad 4 a+2-6 c=0$
$\Rightarrow \quad 2 a+1-3 c=0$....(ii)
Also, planes $\pi$ and Eq. (i) are perpendicular.
$\Rightarrow \quad a+5-6 c=0$
$\Rightarrow \quad a=6 c-5$...(iii)
On putting $a=6 c-5$ in Eq. (ii),
$2(6 c-5)+1-3 c=0$
$\Rightarrow \quad 12 c-10+1-3 c=0$
$\Rightarrow \quad 9 c=9$
$\Rightarrow \quad c=1$
On putting $c=1$ in Eq. (iii),
$a=6-5=1$
Thus, $a=c=1$
On putting $a=c=1$ in Eq. (i),
$x+y+z-d=0$
This plane passes through $(-1,2,5)$
$\begin{array}{ll}\Rightarrow & -1+2+5-d=0 \\ \Rightarrow & d=6\end{array}$
Now, $3(a+c)$
$=3(1+1)=3 \times 2=6$
$\therefore \quad 3(a+c)=d$