Let $$Q$$ is the point of intersection of lines

$$\begin{aligned} & \mathrm{L}_1: \frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}=\lambda \text { and } \\ & \mathrm{L}_2: \frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}=\mu \\ & \mathrm{Q}=(1+2 \lambda,-\lambda,-3+\lambda) \\ & =(4+\mu,-3+\mu,-3+2 \mu) \\ & \Rightarrow \lambda=3-\mu \text { and } \lambda=2 \mu \\ & \Rightarrow \lambda=2 \mu=3-\mu \\ & \Rightarrow \mu=1, \lambda=2 \\ & \therefore \quad Q=(5,-2,-1) \\ \end{aligned}$$

Given, the plane $$a x+b y+c z=d$$ passing through the intersection of lines $$\mathrm{L}_1$$ and $$\mathrm{L}_2$$

$$\therefore \quad \mathrm{Q}(5,-2,-1)$$ satisfy the equation

$$\begin{aligned} & a x+b y+c z=d \\ \Rightarrow & 5 a-2 b-c=d \quad \text{... (i)} \end{aligned}$$

Given, the plane $$a x+b y+c z=d$$ is perpendicular to both the planes $$3 x+5 y-6 z =4$$ and $$7 x+y+2 z=3$$

$$\therefore \quad a \hat{i}+b \hat{j}+c \hat{k}$$ is parallel to

$$\begin{aligned} & (7 \hat{i}+\hat{j}+2 \hat{k}) \times(3 \hat{i}+5 \hat{j}-6 \hat{k}) \\ \Rightarrow & a \hat{i}+b \hat{j}+c \hat{k}=\gamma\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 7 & 1 & 2 \\ 3 & 5 & -6 \end{array}\right| \\ \Rightarrow & a \hat{i}+b \hat{j}+c \hat{k}=-16 \gamma \hat{i}+48 \gamma \hat{j}+32 \gamma \hat{k} \\ \Rightarrow & a=-16 \gamma, b=48 \gamma, c=32 \gamma \end{aligned}$$

If $$\gamma=\frac{-1}{16}$$

$$\Rightarrow a=1, b=-3, c=-2$$

Put $$a=1, b=-3$$ and $$c=-2$$ in the equation (i)

$$\begin{aligned} & \Rightarrow d=5+6+2 \\ & \Rightarrow d=13 \end{aligned}$$

Hints:

(i) If a plane $$\mathrm{P}=0$$ is perpendicular to both planes $$P_1=0$$ and $$P_2=0$$, then normal vector of $$P$$ is parallel to cross Product of normals of planes $$\mathrm{P}_1$$ and $$\mathrm{P}_2$$.

(ii) Recall the method of finding the point of intersection of two lines in three Dimensional co-ordinate geometry.