Answer
$1\left( {x - 1} \right) + 4\left( {y - 0} \right) - 3\left( {z + 2} \right) = 0$
Work Step by Step
We have the vector equation of the plane: $\left( {1,4, - 3} \right)\cdot\left( {x,y,z} \right) = 7$.
$x+4y-3z=7$.
Let ${x_0} = 1$ and ${y_0} = 0$. We may choose a point $P = \left( {{x_0},{y_0},{z_0}} \right)$ such that $P$ satisfies the equation $x+4y-3z=7$. Substituting ${x_0} = 1$ and ${y_0} = 0$ in the equation gives ${z_0} = - 2$. Thus, the point $P = \left( {{x_0},{y_0},{z_0}} \right) = \left( {1,0, - 2} \right)$ is on the plane.
By Theorem 1 of Section 13.5, we have $a=1$, $b=4$, $c=-3$. Therefore, the equation of the plane in scalar form is
$a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0$
$1\left( {x - 1} \right) + 4\left( {y - 0} \right) - 3\left( {z + 2} \right) = 0$
