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

Answer: \(\bar r\cdot(\hat j - \hat k) = -1\)

\(\overline{AB} = -\hat i + \hat j + \hat k\), \(\overline{AC} = 3\hat i + 4\hat j + 4\hat k\).

Normal \(\bar n = \overline{AB} \times \overline{AC} = \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & 1 \\ 3 & 4 & 4 \end{vmatrix} = \hat i(4 - 4) - \hat j(-4 - 3) + \hat k(-4 - 3) = 7\hat j - 7\hat k\).

Take \(\bar n = \hat j - \hat k\). \(\bar a\cdot\bar n = (\hat i + \hat j + 2\hat k)\cdot(\hat j - \hat k) = 1 - 2 = -1\).

Vector equation: \(\bar r\cdot(\hat j - \hat k) = -1\) (cartesian form \(y - z + 1 = 0\); B and C satisfy it: \(2 - 3 = -1\), \(5 - 6 = -1\)).