#### Explain solution RD Sharma class 12 chapter Determinants exercise 5.3 question 5 maths

$\boldsymbol{Answer\! :} \lambda =-5$

$\boldsymbol{H\! int\! :} I\! f \left ( 1,-5 \right )\! , \left ( -4,5 \right )\; and\; \left ( \lambda ,7 \right )$

$\boldsymbol{Explanation\! :} V\! ertices\; are \left ( 1,-5 \right )\! , \left ( -4,5 \right )\; and\; \left ( \lambda ,7 \right )$

$\text { Determinant }=\left|\begin{array}{ccc} 1 & -5 & 1 \\ -4 & 5 & 1 \\ \lambda & 7 & 1 \end{array}\right|=0$

$\Rightarrow 1\left|\begin{array}{ll} 5 & 1 \\ 7 & 1 \end{array}\right|-(-5)\left|\begin{array}{cc} -4 & 1 \\ \lambda & 1 \end{array}\right|+1\left|\begin{array}{cc} -4 & 5 \\ \lambda & 7 \end{array}\right|=0$

$\Rightarrow 1(5-7)+5(-4-\lambda)+1(-28-5 \lambda)=0$

$\Rightarrow -2-20-5\lambda -28-5\lambda =0$

$\Rightarrow -10\lambda -50=0$

$\Rightarrow -10\lambda =50$

$\Rightarrow \lambda =\frac{50}{-10}$

$\Rightarrow \lambda =-5$

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