Equation of plane passing through $(2,-3,4)$ having $\operatorname{DR}^{\prime} s(a, b, c)$ is given by
$$
a \cdot(x-2)+b(y+3)+c(z-4)=0
$$


Since, Eq. (i) is perpendicular to both the planes $2 x-3 y+5 z=2$ and $x+y+2 z=3$

On solving Eqs. (i) and (ii), we get
$$
\frac{a}{-6-5}=\frac{b}{5-4}=\frac{c}{2+3}=k \text { (say) }
$$
i.e. $\quad(a, b, c)=(-11 k, k, 5 k)$
From Eq. (i), we get
$\begin{array}{cr} & -11 x+y+5 z-2 \times(-11)+3 \times 1-4 \times 5=0 \\ \Rightarrow & -11 x+y+5 z=-5 \\ \Rightarrow & x-\frac{1}{11} y-\frac{5}{11} z=\frac{5}{11}\end{array}$
On comparing with $x+p y+q z=r, r=5 / 11$