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The length of the perpendicular drawn from the point (2,1,4) to
the plane containing the lines
r→=(i^+j^)+λ(i^+2j^−k^) and r→=(i^+j^)+μ(−i^+j^−2k^) is \overset{\rightarrow}{r} = (\overset{\hat{}}{i} + \overset{\hat{}}{j}) + \lambda(\overset{\hat{}}{i} + 2\overset{\hat{}}{j} - \overset{\hat{}}{k})\ and\ \overset{\rightarrow}{r} = (\overset{\hat{}}{i} + \overset{\hat{}}{j}) + \mu( - \overset{\hat{}}{i} + \overset{\hat{}}{j} - 2\overset{\hat{}}{k})\ is\:

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Normal vector of the plane is
$\left| \begin{matrix}
\overset{\hat{}}{i} & \overset{\hat{}}{j} & \overset{\hat{}}{k} \\
1 & 2 & - 1 \\
- 1 & 1 & - 2
\end{matrix} \right| =$
−3i^+3j^+3k^- 3\overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 3\overset{\hat{}}{k}
 now equation of plane \ now\ equation\ of\ plane\
−3x+3y+3z=c- 3x + 3y + 3z = c
(1,1,0) will satisfy the plane (1,1,0)\ will\ satisfy\ the\ plane\
⇒−3+3+0=c\Rightarrow - 3 + 3 + 0 = c
⇒c=0\Rightarrow c = 0
⇒−3x+3y+33=0\Rightarrow - 3x + 3y + 33 = 0
Distance from (2, 1, 4)
=∣−6+3+1227∣=∣933∣=3 units = \left| \frac{- 6 + 3 + 12}{\sqrt{27}} \right| = \left| \frac{9}{3\sqrt{3}} \right| = \sqrt{3}\ units\

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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