JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 5)
The distance of the point $$(1, 1, 1)$$ from the plane passing through the point $$(-1, -2, -1)$$ and whose normal is perpendicular to both the lines $${L_1}$$ and $${L_2}$$ is :
$${2 \over {\sqrt {75} }}$$
$${7 \over {\sqrt {75} }}$$
$${13 \over {\sqrt {75} }}$$
$${23 \over {\sqrt {75} }}$$
Explanation
The equation of the plane passing through the point ($$-1,-2,-1$$) and whose normal is perpendicular to both the given lines L$$_1$$ and L$$_2$$ written as
$$(x + 1) + 7(y + 2) - 5(z + 1) = 0$$
i.e., $$x + 7y - 5z + 10 = 0$$
The distance of the point (1, 1, 1) from the plane
$$ = \left| {{{1 + 7 - 5 + 10} \over {\sqrt {1 + 49 + 25} }}} \right|$$
$$ = {{13} \over {\sqrt {75} }}$$ units
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