Question

Find the vector equation of the plane determined by the points (1, 0, −1),
(0, 1, 1) and (−1, 1, 0). Alsofind the point of intersection of the line

Answer 1

$points \\P(1, 0, −1), Q(0, 1, 1)\ and\ R(−1, 1, 0)$

The equation of a plane passing through three points

here are finding the cross product of $vector \ QP \&vector RP$ which gives us the direction of the normal to the plane

$\begin{vmatrix} x-x_1& y-y_1&z-z_1 \\ x_2-x_1 & y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1 \end{vmatrix}$

$\begin{vmatrix} x-1 & y-0&z+1 \\ 0-1 & 1-0 &1-(-1)\\-1-1&1-0&0-(-1)\end{vmatrix}$

$x−1−z−1−4y+2z+2+y−2x+2=0\\−x−3y+z+2=0$

Equation of plane is $:x+3y-z=2$