If L 1 is the line of intersection of the planes 2x – 2y + 3z – 2 = 0, x – y + z + 1 = 0 and L 2 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 L 1 and L 2 , is :
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L 1 : 
Let a point on L 1 (0, 5, 4) and dr; s of L 1 be a, b, c
2a 1 + 2b 1 + 3c 1 = 0
a 1 + b 1 + c 1 = 0
so dr's of L 2 be a 2 , b 2 , c 2
dr's of L 2 can be 3, –5, –7
so dr's of normal to the plane ca be a + b + oc = 0
3a – 5b –7c = 0
equation req. plane 7x – 7 (y –5) + 8 (z–4) = 0
7x –7y + 8z + 3 = 0
so req. distance = 
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