JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
If the system of linear equations $8x+y+4z=-2$, $x+y+z=0$, $\lambda x-3y=\mu$ has infinitely many solutions, then the distance of the point $\left(\lambda,\mu,-\frac12\right)$ from the plane $8x+y+4z+2=0$ is :
Determinant
Solution
Homogeneity: $\begin{vmatrix}8&1&4\\1&1&1\\\lambda&-3&0\end{vmatrix}=0\Rightarrow 3\cdot(-4)\ldots=0$ gives $\lambda=4$. Then $\begin{vmatrix}8&1&-2\\1&1&0\\4&-3&\mu\end{vmatrix}=0\Rightarrow\mu=-2$. Point $(4,-2,-\tfrac12)$: $8(4)-2+4(-\tfrac12)+2=30$; $\sqrt{64+1+16}=9$. Distance $=\dfrac{30}{9}=\dfrac{10}{3}$. Option (D).