NIMCET 2017 — Mathematics PYQ

Concept:

Let $\vec{a}=a_1\vec{i}+b_1\vec{j}+c_1\vec{k}$, $\vec{b}=a_2\vec{i}+b_2\vec{j}+c_2\vec{k}$ and $\vec{c}=a_3\vec{i}+b_3\vec{j}+c_3\vec{k}$ be the three vectors

Condition for coplanarity: $\vec{a}\cdot (\vec{b}\times \vec{c})=\left|\begin{array}{ccc}{a}_1& amp;b_1& amp;c_1\\ a_2& amp;b_2& amp;c_2\\ a_3& amp;b_3& amp;c_3\end{array}\right|=0$

Calculation:
Here, $\vec{a}\hat{ı}+\vec{j}+\vec{k}$, $\vec{b}\hat{ı}+\vec{j}+\vec{k}$, $\vec{i}+\vec{j}+\vec{c}\hat{k}$ ($a\ne b\ne c\ne 1$) are co-planar
$$\left|\begin{array}{ccc}a& amp;1& amp;1\\ 1& amp;b& amp;1\\ 1& amp;1& amp;c\end{array}\right|=0$$
$c_1\to c_1-c_2$, $c_2\to c_2-c_3$
$$</span><spanstyle="font-size:14pt;">\left|\begin{array}{ccc}a-1& amp;0& amp;1\\ 1-b& amp;b-1& amp;1\\ 0& amp;1-c& amp;c\end{array}\right|=0$$
$c(a-1)(b-1)+(1-b)(1-c)=0$
$$\frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=0$$
$$</span><spanstyle="font-size:14pt;">\frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1$$
$$\frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1$$

Hence, option (4) is correct.