The distance of the plane 3x+4y+5z+19=0 from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

The distance of the plane $3 x + 4 y + 5 z + 19 = 0$ from the point (1,-1,1) measured along a line parallel to the line with direction ratios 2,3,1 is

A
$\frac{23}{5 \sqrt{2}}$
B
$\frac{\sqrt{71}}{5 \sqrt{2}}$
C
$\sqrt{14}$
D
$\sqrt{23}$

Solution:

Line passing through (1,-1,1) is $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{1} = t$

$P (2 t + 1, 3 t - 1, t + 1)$ lies on $3 x + 4 y + 5 z + 19 = 0$

$t = - 1 \Rightarrow P (- 1, - 4, 0) Q (1, - 1, 1)$

$P Q = \sqrt{14}$

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