Let the equation of the plane passing through the line x - 2y - z - 5 = 0 = x + y + 3z - 5 and parallel to the line x + y + 2z - 7 = 0 = 2x + 3y + z - 2 be ax + by + cz = 65. Then the distance of the point (a, b, c) from the plane 2x + 2y - z + 16

Let the equation of the plane is (x-2 y-z-5)+(x+y+3 z-5)=0 it's parallel to the line x+y+2 z-7=0=2 x+3 y+z-2 So, vector along the line |arrayllli & j & k \\ 1 & 1 & 2 \\ 2 & 3 & 1array| =-5 i+3 j+k Plane is parallel to line -5(1+)+3(-2+)+1(-1+3 )=0 =12

Let the equation of the plane passing through the line x 2y : null (Mathematics)

Let the equation of the plane is

(x-2 y-z-5)+\lambda(x+y+3 z-5)=0

\because \quad

it's parallel to the line

x+y+2 z-7=0=2 x+3 y+z-2

So, vector along the line

\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|

=-5 \hat{i}+3 \hat{j}+\hat{k}

\because

Plane is parallel to line

\therefore -5(1+\lambda)+3(-2+\lambda)+1(-1+3 \lambda)=0

\lambda=12

So, by (i)

13 x+10 y+35 z=65

\therefore a=13, b=10, c=35

and

d=\frac{26+20-35+16}{\sqrt{9}}=9

Explanation

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