The given planes are $\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7$ and $\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$ $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k},$ we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7$ and $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$ i.e. 2x + 2y – 3z – 7 = 0 ...(1) and 2x + 5y + 3z – 9 = 0 ...(2) Any plane through the intersection of (1) and (2) is : (2x + 2y – 3z – 7) + k(2x + 5y + 3z – 9) = 0 ...(3) Since it passes through (2, 1, 3). ∴ (4 + 2 – 9 – 7) + k (4 + 5 + 9 – 9) = 0 $\Rightarrow -10 + k \times 9 = 0$ $\Rightarrow k = \frac{10}{9}$ Putting in (3), we get (2x + 2y – 3z – 7) + $\frac{10}{9}$ (2x + 5y + 3z – 9) = 0 ⇒ 18x + 18y – 27z – 63 + 20x + 50y + 30z – 90 = 0 ⇒ 38x + 68y + 3z – 153 = 0$\Rightarrow \vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153$which is the required equation.