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(i)
Since $|a| = |b|$

$\Rightarrow \sqrt{(2p)^2+(3p)^2+(6p)^2} = \sqrt{1+2^2+2^2}$

$49p^2 = 9$

$p = \frac{3}{7}$ since $p > 0$

(ii)
$(a+b) \bullet (a-b)$

$= a \bullet a + b \bullet a - a \bullet b - b \bullet b$

$= |a|^2 - |b|^2$

$= 0$ since $|a| = |b|$