Q100 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

A plane $\pi$ passes through the points $(5,1,2),(3,-4,6$ and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosin of a normal to the plane $\pi$, then $|3 l+2 m+5 n|=$

  • A
    $3 p$
  • B
    $2 p$
  • Correct answer
    C
    $p$
  • D
    $\frac{p}{2}$

Solution

$P(5,1,2), Q(3,-4,6), R(7,0,-1)$ Equation of $P$ line passing through $P Q R$ is $\begin{aligned} & \left|\begin{array}{ccc} x-5 & y-1 & z-2 \\ 3-5 & -4-1 & 6-2 \\ 7-5 & 0-1 & -1-2 \end{array}\right|=0 \\ & \Rightarrow|9 x+2 y+12 z-12|=0 \end{aligned}$ Direction cosines of normal to plane $\pi$ are $\begin{aligned} & (l, m, n)=\left(\frac{19}{\sqrt{509}}, \frac{2}{\sqrt{509}}, \frac{12}{\sqrt{509}}\right) \\ & p=\frac{|0+0+0+121|}{\sqrt{509}}=\frac{121}{\sqrt{509}} \\ & |3 l+2 n+5 m|=\frac{57+4+60}{\sqrt{509}}=\frac{121}{\sqrt{509}}=p . \end{aligned}$

Hence, the correct answer is option: 'C'.

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