Vector Form : Here $\vec{a}=\hat{i}+\hat{4j}+\hat{6k}$ and $\hat{n}=\hat{i}-\hat{2j}+\hat{k}$ The equation of the plane is : $(\vec{r}-\vec{a})\cdot\vec{n}=0$ Cartesian Form :If (a, b, c) are direction ratios of the normal to the plane, then the equation of the plane through $(x_1, y_1, z_1)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$This passes through (1, 4, 6)$\therefore x_1 = 1, y_1 = 4, z_1 = 6$Direction ratios of normal $\hat{i}-\hat{2j}+\hat{k}$ are < 1, –2, 1 >$\therefore a = 1, b =-2, c = 1$∴ The equation of the plane is :1 (x – 1) – 2 (y – 4) + 1 (z – 6) = 0⇒ x – 2y + z – 1 + 8 – 6 = 0 ⇒ x – 2y + z + 1 = 0