Calculus 3 problem help

Find an equation for the plane that passes through the line of intersection of the two planes -10*x+6*y+4*z = 6 and 8*x - 6*y + 8*z = 4, and is perpendicular to the plane 8*x + y + 2*z = -8

any help is very much appreciated

First we find line of intersection of the two planes. To do this we need a point on the line (i.e. on both planes) and direction vector for line.

Let x = 1. Then equations of planes become:
6y + 4z = 16
−6y + 8z = −4
------------------- add
0y + 12z = 12
z = 1

6y + 4z = 16
6y + 4 = 16
6y = 12
y = 2

So point that is on both planes (and therefore on line of intersection)
is (1, 2, 1)

Vector < −10, 6, 4 > is normal to first plane
Vector < 8, −6, 8 > is normal to second plane
Their cross product will produce vector that is orthogonal to both vectors, and therefore parallel to both planes:

< −10, 6, 4 > x < 8, −6, 8 > = < 72, 112, 12 > = 4 < 18, 28, 3 >

Therefore line has direction vector < 18, 28, 3 > and parametric equation:
x = 1 + 18k
y = 2 + 28k
z = 1 + 3k

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Now we find equation of plane that passes through line above, and is perpendicular to the plane 8x + y + 2z = −8

Vector < 8, 1, 2 > is normal to given plane. Therefore, it is parallel to plane that we are looking for.
Direction vector of line < 18, 28, 3 > is also parallel to plane we are looking for.
Their cross product will produce vector that is orthogonal to both vectors, and therefore orthogonal to the plane we are looking for:

< 18, 28, 3 > x < 8, 1, 2 > = < 53, −12, −206 >

So plane has normal < 53, −12, −206 > and passes through point (1, 2, 1)

Equation of plane:
53(x−1) − 12(y−2) − 206(z−1) = 0
53x − 53 − 12y + 24 − 206z + 206 = 0
53x − 12y − 206z + 177 = 0

53x − 12y − 206z = −177

Mαthmφm

the answer available in calculus book.