The equation of the plane which is at a distance of 3 units from the origin and whose normal has the d.c’s (13, 13, 13) is

p=3 (ℓ,m,n)=(13,13,23) Equation of the plane is ℓx+my+nz=p 13x+13y+13z=3 x + y + z = 3

The equation of the plane which is at a distance of 3 units from the origin and whose normal has the d.c’s (13, 13, 13) is

p=3 (ℓ,m,n)=(13,13,23) Equation of the plane is ℓx+my+nz=p 13x+13y+13z=3 x + y + z = 3

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The equation of the plane which is at a distance of $\sqrt{3}$ units from the origin and whose normal has the d.c’s $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ is

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x + y + z = 3

x – y + z = 6

3x–12y + 4z = 26

x + y – z = 3

answer is A.

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$p = \sqrt{3}$ $(ℓ, m, n) = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}})$

Equation of the plane is

$ℓ x + m y + n z = p$

$\frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = \sqrt{3}$

x + y + z = 3

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The equation of the plane which is at a distance of 3 units from the origin and whose normal has the d.c’s (13, 13, 13) is

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The equation of the plane which is at a distance of $\sqrt{3}$ units from the origin and whose normal has the d.c’s $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$ is