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(B) The equation of a plane,when the direction ratios $(a, b, c)$ of the normal and the length of the perpendicular $p$ from the origin are given,is $

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$-3x + 2y + 6z - 49 = 0$

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(B) The equation of a plane,when the direction ratios $(a, b, c)$ of the normal and the length of the perpendicular $p$ from the origin are given,is $ax + by + cz = p \sqrt{a^2 + b^2 + c^2}$.Given,$(a, b, c) = (-3, 2, 6)$ and $p = 7$.Substituting these values into the formula:$-3x + 2y + 6z = 7 \sqrt{(-3)^2 + 2^2 + 6^2}$$-3x + 2y + 6z = 7 \sqrt{9 + 4 + 36}$$-3x + 2y + 6z = 7 \sqrt{49}$$-3x + 2y + 6z = 7 	imes 7$$-3x + 2y + 6z = 49$Thus,the equation is $-3x + 2y + 6z - 49 = 0$.

If the length of the perpendicular drawn from the origin to a plane is $7$ units and its direction ratios are $-3, 2, 6$,then the equation of the plane is:

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