If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is th

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If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y - z - 3 =0, 3x - y + 2z - 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2 is :

Accepted Answer

Plane passes through line of intersectuion of first two planes is
(2x - 2y + 3z - 2) + $λ$ (x - y + z + 1) = 0
x( $λ$ + 2) - y(2 + $λ$ ) + z( $λ$ + 3) + ( $λ$ - 2) = 0 ..... (1)
is having infinite number of solution with
x + 2y - z - 3 = 0 and 3x - y + 2z - 1 = 0 then
$|\begin{matrix} (λ + 2) & - (λ + 2) & (λ + 3) \\ 1 & 2 & - 1 \\ 3 & - 1 & 2 \end{matrix}|$ = 0
Solving $λ$ = 5
7x - 7y + 8z + 3 = 0
perpendicular distance from (0, 0, 0)
is $\frac{3}{\sqrt{162}} = \frac{1}{3 \sqrt{2}}$

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