Consider points A, B, C and D with position vectors $7\hat i-4\hat j+7\hat k, \hat i-6\hat j+10\hat k, -\hat i - 3\hat j + 4\hat k$ and $5\hat i-\hat j+\hat k$ respectively. Then, ABCD is a

rhombus

We have,

$\vec{AB}=-6\hat i-2\hat j + 3\hat k, \vec{BC}=-2\hat i +3\hat j-6k$

$\vec{CD}=6\hat i+2\hat j - 3\hat k, \vec{DA}=2\hat i -3\hat j+6k$

$\vec{AC}=-8\hat i+\hat j-3\hat k$ and $\vec{BD}=4\hat i+5\hat j-9\hat k$

Clearly, $|\vec{AB}|=| \vec{BC}|=|\vec{CD}|=|\vec{DA}|$ and $\vec{AC}. \vec{BD} = 0$

Hence, ABCD is a rhombus.