Find the vector equation of the plane passing through the points \(A(2, 1, 1)\), \(B(0, 2, 3)\) and \(C(4, 5, 6)\).

Answer: \(\bar{r}\cdot(3\hat{i} - 14\hat{j} + 10\hat{k}) = 2\)

\(\overline{AB} = -2\hat{i} + \hat{j} + 2\hat{k}\), \(\overline{AC} = 2\hat{i} + 4\hat{j} + 5\hat{k}\).

Normal \(\bar{n} = \overline{AB}\times\overline{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 2 \\ 2 & 4 & 5 \end{vmatrix} = \hat{i}(5 - 8) - \hat{j}(-10 - 4) + \hat{k}(-8 - 2) = -3\hat{i} + 14\hat{j} - 10\hat{k}\).

Equation: \(\bar{r}\cdot\bar{n} = \bar{a}\cdot\bar{n}\), where \(\bar{a}\cdot\bar{n} = 2(-3) + 1(14) + 1(-10) = -2\).

So \(\bar{r}\cdot(-3\hat{i} + 14\hat{j} - 10\hat{k}) = -2\), i.e. \(\bar{r}\cdot(3\hat{i} - 14\hat{j} + 10\hat{k}) = 2\). (Check: B gives \(0 - 28 + 30 = 2\), C gives \(12 - 70 + 60 = 2\).)

Equivalently, in parametric form \(\bar{r} = \bar{a} + \lambda(\bar{b} - \bar{a}) + \mu(\bar{c} - \bar{a})\): \(\bar{r} = (2\hat{i} + \hat{j} + \hat{k}) + \lambda(-2\hat{i} + \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} + 5\hat{k})\).