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Question

# The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 2√3 from the point (3, 1,−1) is

A
5x11y+z=17
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B
2x+y=321
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C
x+y+z=3
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D
x2y=12
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Solution

## The correct option is A 5x−11y+z=17Equation of the required plane isL=(x+2y+3x−2)+K(x−y+z−3)=0⇒(1+K)x+(2−K)y+(3+K)z−(2+3K)=0Its distance from (3,1,−1) is 2√3⇒∣3(1+K)+(2−K)−(3+K)−(2+3K)∣√(K+1)2+(2−K)2+(3+K)2 =2√3⇒43=(−2K)22K2+4K−14⇒3K2+4K+14=3K2⇒K=−72⇒−52x+112y−z2+172=0⇒−5x+11y−z+17=0⇒5x−11y+z=17

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