JEE 2023 — Mathematics PYQ

Let $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

Now, $\vec{a}\cdot \vec{c}=-12$
$$\Rightarrow 6c_1+9c_2+12c_3=-12\dots (i)$$

Also, $\vec{c}\cdot (\hat{i}-2\hat{j}+\hat{k})=5$
$$\Rightarrow c_1-2c_2+c_3=5\dots (ii)$$

Now, $\vec{a}\times \vec{c}=\vec{a}\times \vec{b}$
$\Rightarrow \vec{a}\times (\vec{c}-\vec{b})=0$
$\Rightarrow \vec{a}$ is parallel to $(\vec{c}-\vec{b})$
$\Rightarrow \vec{a}=\lambda (\vec{c}-\vec{b})$
$\Rightarrow 6\hat{i}+9\hat{j}+12\hat{k}=\lambda (c_1-\alpha )\hat{i}+\lambda (c_2-11)\hat{j}+\lambda (c_3+2)\hat{k}$

On comparing, we get
$$c_1=\frac{6}{\lambda }+\alpha ,c_2=\frac{9}{\lambda }+11,c_3=\frac{12}{\lambda }-2$$

Put these values in (ii), we get
$$\frac{6}{\lambda }+\alpha -\frac{18}{\lambda }-22+\frac{12}{\lambda }-2=5$$
$\Rightarrow \alpha =29$

From (i) and values of $c_1,c_2,c_3$, and $\alpha$ we have
$$6\left(\frac{6}{\lambda }+29\right)+9\left(\frac{9}{\lambda }+11\right)+12\left(\frac{12}{\lambda }-2\right)=-12$$
$\Rightarrow \frac{261}{\lambda }=-261\Rightarrow \lambda =-1$

So, $c_1=23,c_2=2,c_3=-14$

$$\therefore \vec{c}\cdot (\hat{i}+\hat{j}+\hat{k})=(23\hat{i}+2\hat{j}-14\hat{k})\cdot (\hat{i}+\hat{j}+\hat{k})$$
$$=23+2-14=11$$