Question

Find the unit vector perpendicular to the plane ABC where the position vectors A, B and C are 2 i - j + k, i + j + 2 k and 2 i + 3 k.

Answer 1

$2i - j + k + i + j + 2k + 2i + 3k$
$= 5i + 6k$
$\sqrt{(5) { }^{2} + (6) {}^{2} } = \sqrt{25 + 36 } = \sqrt{61}$
then answer is =
$\frac{5i + 6k}{ \sqrt{61} }$

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