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Question

Let p1:x2y+3z=5 and p2:2x+3y+z+4=0 be two planes. If P is the foot of the perpendicular dropped from the origin O to the line of intersection of the planes, then

A
OP((^i2^j+3^k)×(2^i+3^j+^k))=0
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B
if Q is another point on the line of intersection of planes, then equation of the plane containing a triangle OPQ is 14x+7y+17z=0.
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C
equation of the plane perpendicular to the line of intersection of the planes and passing through (1,1,1) is 11x5y7z+1=0.
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D
If N1 and N2 are foot of the perpendicular from the origin O to the planes p1 and p2 respectively, then ON1+ON2 is equal to 914.
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Solution

The correct option is D If N1 and N2 are foot of the perpendicular from the origin O to the planes p1 and p2 respectively, then ON1+ON2 is equal to 914.
Let r=x^i+y^j+z^k
p1r(^i2^j+3^k)=5p2r(2^i+3^j+^k)=4


DR's of intersection line (L) of p1 and p2 is
∣ ∣ ∣^i^j^k123231∣ ∣ ∣=11^i+5^j+7^k

Now, OP is to L
and ^i2^j+3^k, 2^i+3^j+^k are also perpendicular to L.
This means OP((^i2^j+3^k)×(2^i+3^j+^k))=0

Equation of planes passing through intersection of p1 and p2 is
p1+λp2=0(x2y+3z5)+λ(2x+3y+z+4)=0x(1+2λ)+y(3λ2)+z(3+λ)5+4λ=0 (1)
Put (0,0,0)
0+0+0+4λ5=0λ=54
Putting this value in equation (1),
x(1+2(54))+y(3(54)2)+z(3+54)+0=014x+7y+17z=0

Vector along the line of intersection of planes is
∣ ∣ ∣^i^j^k123231∣ ∣ ∣=11^i+5^j+7^k

11(x1)+5(y1)+7(z1)=0
or, 11x5y7z+1=0

ON1+ON2∣ ∣512+22+32∣ ∣+∣ ∣422+32+12∣ ∣=514+414=914

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