The given planes are $\vec{r}\cdot(\hat{i}+\hat{2j}+\hat{3k})-4=0$ , $\vec{r}\cdot(\hat{2i}+ \hat{j}- \hat{k})+5=0$ and $\vec{r}\cdot(\hat{5i}+\hat{3j}-\hat{6k})+8=0$ Putting $\vec{r}= \hat{xi}+\hat{yj}+\hat{zk}$, we get x + 2y + 3z – 4 = 0 ...(1)2x + y – z + 5 = 0 ...(2)and 5x + 3y – 6z + 8 = 0 ...(3)Any plane through the intersection of (1) and (2) is(x + 2y + 3z – 4) + k (2x + y – z + 5) = 0⇒ (1 + 2k)x + (2 + k) y + (3 – k)z + (–4 + 5k) = 0 ...(4)Now plane (4) is perpendicular to plane (3), we get∴ 5 (1 + 2k) + 3 (2 + k) – 6 (3 – k) = 0⇒ 5 + 10k + 6 + 3k – 18 + 6k = 0⇒ 19k – 7 = 0 ⇒ k = 7/19Putting in (4), we get, 33x + 45y + 50z – 41 = 0,which is the required equation of the plane.