उत्तर (38)

समाधान

मान लीजिए $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$

$(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$

$(5 \hat{i}+\hat{j}+4 \hat{k}) \times \vec{c}=2(7 \hat{i}-7 \hat{j}-7 \hat{k})+24 \hat{j}-6 \hat{k}$

$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z\end{array}\right|=14 \hat{i}+10 \hat{j}-20 \hat{k}$

$\Rightarrow \hat{i}(z-4 y)-\hat{j}(5 z-4 x)+\hat{k}(5 y-x)=14 \hat{i}+10 \hat{j}-20 \hat{k}$

$z-4 y=14,4 x-5 z=10,5 y-x=-20$

$(a-b+i) \cdot \overrightarrow{c}=-3$

$(2 \hat{i}+3 \hat{j}-2 \hat{k}) \cdot \overrightarrow{c}=-3$

$2 x+3 y-2 z=-3$

$\therefore x=5, y=-3, z=2$

$|\overrightarrow{c}|^{2}=25+9+4=38$