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
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L1 is line of the intersection of the plane
2x−2y+3z−2=0 and x−y+z+1=0 and L2 is line
of intersection of the plane x+2y−z−3=0 and 3x−y+2z−1=0
Since L1 is parallel to ∣∣
∣
∣∣^ij^k2−231−11∣∣
∣
∣∣=^i+^j
L2 is parallel to ∣∣
∣
∣∣^i^j^k12−13−12∣∣
∣
∣∣=3^i−5^j−7^k
Also, L2 passes through (57,87,0)
(put z=0 in last two planes)
so, equation of plane is
∣∣
∣
∣
∣∣x−57y−87z1103−5−7∣∣
∣
∣
∣∣=0
7x−7y+8z+3=0
Now, perpendicular distance from origin is
∣∣
∣∣3√72+72+82∣∣
∣∣=3√162=13√2
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