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If the system of linear equations.

    \mathrm{8 x+y+4 z=-2}

        \mathrm{x+y+z=0}

    \mathrm{\lambda x-3 y=\mu}

has infinitely many solutions, then the distance of the point \mathrm{\left(\lambda, \mu,-\frac{1}{2}\right)} from the plane \mathrm{8 x+y+4 z+2=0 \text { is : }}

Option: 1

\mathrm{3 \sqrt{5}}


Option: 2

4


Option: 3

\frac{26}{9}


Option: 4

\frac{10}{3}


Answers (1)

best_answer

\mathrm{D=\left|\begin{array}{rrr} 8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & -3 & 0 \end{array}\right|=0 \Rightarrow \lambda=4}

\mathrm{Also\; D_{1}=D_{2}=D_{3}=0}

\mathrm{So\;\mu=-2 }\\

\mathrm{\text { point }\left(4,-2,-\frac{1}{2}\right)}

\mathrm{Distance \;from\; plane =\left(\frac{10}{3}\right)}

Hence correct option is 4

Posted by

Rakesh

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