{aligned} $$

$$ \begin{aligned} & \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(-6)}{6}=2 \\\\ & \Rightarrow x=5, y=0, z=5 \end{aligned} $$

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 111 English Explanation

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 111 English Explanation

Now, $\overrightarrow {PQ} =4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and

$\overrightarrow{P R}=5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$\therefore$ Area of the $\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|$

Now, $\overrightarrow{P Q} \times \overrightarrow{P R}$

$$ \begin{aligned} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & -2 & 2 \\ 5 & 8 & 4 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(-8-16)-\hat{\mathbf{j}}(16-10)+\hat{\mathbf{k}}(32+10) \\\\ & =-24 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}} \end{aligned} $$

$$ \begin{aligned} \therefore \frac{1}{2} \mid(-24 \hat{\mathbf{i}} & -6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}})|=|-12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+21 \hat{\mathbf{k}} \mid \\\\ & =\sqrt{(-12)^2+(-3)^2+(21)^2} \\\\ & =\sqrt{144+9+441}=\sqrt{594} \end{aligned} $$

$\therefore$ The square of the area of the $\triangle P Q R=594$

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