The area of the quadrilateral ABCD with vertices A ( 2, 1, 1 ) , B...

Given,

The quadrilateral $ABCD$ with vertices $A ( 2, 1, 1 ) , B ( 1, 2, 5 ) , C ( - 2, - 3, 5 )$ and $D ( 1, - 6, - 7 )$

Now finding, \overset{\to}{AB} = - \overset{^}{i} + \overset{^}{j} + 4\overset{^}{k} and \overset{\to}{AD} = - \overset{^}{i} - 7\overset{^}{j} - 8\overset{^}{k}

And plotting the diagram we get,

Now we know that,

Area of $∆ ABD$ is given by,

= \dfrac{1}{2}\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 1 & 1 & 4 \\ - 1 & - 7 & 8\end{cases}\right|

= | 10\overset{^}{i} - 6\overset{^}{j} + 4\overset{^}{k} | = 2\sqrt{38}

Now finding, \overset{\to}{CB} = 3\overset{^}{i} + 5\overset{^}{j} and \overset{\to}{CD} = 3\overset{^}{i} - 3\overset{^}{j} - 12\overset{^}{k}
So, Area of ∆ CBD = \dfrac{1}{2}\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 3 & - 3 & - 12 \\ 3 & 5 & 0\end{cases}\right|

= | 6 ( 5\overset{^}{i} - 3\overset{^}{j} - 2\overset{^}{k} ) |

$= 6\sqrt{38}$
$∴$ Area of quadrilateral $ABCD = 8\sqrt{38}$ square units