Answer
(a) $\lt 4,-3,4 \gt$
(b) $\frac{\sqrt {41}}{2}$
Work Step by Step
(a) $AB= \lt 1,0,-1 \gt$ and $AC= \lt 0,4,3 \gt$
The cross product of the two vectors in the plane will be orthogonal to the plane.
$\lt 1,0,-1 \gt \times \lt 0,4,3 \gt= \begin{vmatrix} i&j&k \\1&0&-1& \\0&4&3 \end{vmatrix}$
$=\lt 4,-3,4 \gt$
(b) $Area \triangle ABC=\frac{1}{2}|A|$
$a=AB= \lt 1,0,-1 \gt$ and $b=AC= \lt 0,4,3 \gt$
Here, $A= a \times b= \lt 1,0,-1 \gt \times \lt 0,4,3 \gt$
$= \begin{vmatrix} i&j&k \\1&0&-1& \\0&4&3 \end{vmatrix}$
$A=\lt 4,-3,4 \gt$
$|A|=\sqrt {4^2+(-3)^2+4^2}=\sqrt {41}$
Now, $Area \triangle ABC=\frac{1}{2}\sqrt {41}$
$=\frac{\sqrt {41}}{2}$
