Find an equation of the plane.
The plane that passes through the point $ (1, 5, 1) $ and is perpendicular to the planes $ 2x + y - 2z = 2 $ and $ x + 3z = 4 $
$3 x-8 y-z=-38$
In the question they're asking to find the equation of the plane that passes through .151 and is perpendicular to two planes. Given by the equation. To explain why -2 that equal to two and express trees that equal to four. So from the two equations of the plane that is perpendicular to the plane of the given question. We can find the normal vectors from the equations of the perpendicular planes. So the first normal vector is equal to two, I cab plus Jacob minus to kick up. And the second normal vector from the second equation is equal to icap plus three Jacob, it's a zero Jacob Last three. Kick up. So from these two normal vectors you can find them normal vector of the plane asking the question so it will be equal to cross product of these two normal vectors together that is equal to in one cap. Cross into cap that is equal to we will put the values of anyone into in the Stephen 2, -2 and 103. And after computing the cross product of anyone and enter the final value of n. Cap is equal to three. I gap minus a Jacob minus Jacob. And in order to find the equation of the plane. As asking the question, we'll look at another point that is given in the question 15. Fun. And using the normal victor. And the point given the question, we can find the equation of the plane as follows, that is three into X -1 -8 into Y -5 -1 in two, 0 -1 is equal to zero. So the final value of this equation is three x -8 by Miner Said, Is equal to -38. And so this is the equation of the plane as asked from the question.