Question

Find the equation of the plane passing through the intersection of plane
4x - y +z = 10 and
x +y-z = 4 and parallel to the line with direction ratios, 2, 1, 1

Find the perpendicular distance of the point (1,1,1) from this plane.​

Answer 1

The equation of the plane is :

$\boxed{ \bold{3x - 2y + 2z - 6 = 0}}$

The perpendicular distance is :

$\boxed{ \bold{ \frac{3}{ \sqrt{17} } units}}$

For equation of the plane :

• The intersection of two planes is

$p1 + p2 \lambda = 0$

• As it's passes through a point (x1,y1,z1).

• Solve and get the value of lambda and put the value of lambda in p1+p2lambda =0and solve .

For perpendicular distance :

$\boxed{ \bold{l = | \frac{ax1 + by1 + cz1 + d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } }|}}$

solution refer to the attachment

Answer 2

hope this info. will help u ✌✌✌