Answer
(a) $\langle 4,-3,4 \rangle$
(b) $4x-3y+4z=4$
(c) $\frac{\sqrt {41}}{2}$
Work Step by Step
(a) Use the three points given to form two vectors $\vec u=\vec {PQ}=\langle (2-1),(0-0),(-1-0) \rangle=\langle 1,0,-1 \rangle$ and $\vec v=\vec {PR}=\langle (1-1),(4-0),(3-0) \rangle=\langle 0,4,3 \rangle$. The cross product of these two vectors will be perpendicular to the plane: $\vec u\times\vec v=\langle (0\times3+1\times4),(-1\times0-1\times3),(1\times4-0\times0) \rangle=\langle 4,-3,4 \rangle$
(b) The general equation for a plane passing point $P(x_0,y_0,z_0)$ with a surface normal $\vec n=\langle a,b,c \rangle$ is given by $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, thus the equation of the plane can be written as
$4(x-1)-3y+4z=0$ or $4x-3y+4z=4$
(c) The area of the parallelogram formed by vectors $\vec u$ and $\vec v$ is given by $|\vec u\times\vec v|=\sqrt {(4)^2+(-3)^2+(4)^2}=\sqrt {41}$, and the area ($A$) of triangle PQR is half of area of the parallelogram, so that $A=\frac{\sqrt {41}}{2}$
