- A45 squares units
- B55 squares units
- C65 squares units
- Dnone of these
Solution:
Line PA: $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{1}$
Line PB: $\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-6}{1}$
Line PC: $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{-2}$
Then $\text{A}\Big(\frac{7}{2},-\frac{1}{2},\frac{17}{2}\Big)$
$\text{B}\Big(\frac{17}{2},-13,-\frac{3}{2}\Big)$
$\text{C}\Big(-14,\frac{19}{2},21\Big)$
Hence area of $\triangle\text{ABC}=\frac{225\sqrt{14}}{8},$ volume of tetrahedron
$\text{PABC}=\frac{125}{8}\text{cubic units}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$STATEMENT$ $-1: \overline{\mathrm{PQ}} \times(\overline{\mathrm{RS}}+\overline{\mathrm{ST}}) \neq \overrightarrow{0}$. because
$STATEMENT$ $-2: \overline{\mathrm{PQ}} \times \overline{\mathrm{RS}}=\overrightarrow{0}$ and $\overline{\mathrm{PQ}} \times \overline{\mathrm{ST}} \neq \overrightarrow{0}$.
then $\sum_{x \in R }\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right)$ is equal to $........$.