Question

If two distinct points Q, R lie on the line of intersection of the planes – x + 2 y – z = 0 and 3 x – 5 y + 2 z = 0 and
$PQ = PR = \sqrt{18}$
where the point P is (1, –2, 3), then the area of the triangle PQR is equal to

• A. $\frac{2}{3} \sqrt{38}$
• B. $\frac{4}{3}\sqrt{38}$
• C. $\frac{8}{3}\sqrt{38}$
• D. $\sqrt\frac{152}{3}$

Answer 1

Answer: (B)

$\frac{4}{3}\sqrt{38}$

Answer 2

The correct answer is (B) : $\frac{4}{3}\sqrt{38}$

Fig. Triangle

Line L is x = y = z
$\stackrel{→}{PQ}.(\hat{i}+\hat{j}+\hat{k})=0$
$⇒ (α – 3) + α + 2 + α – 1 = 0$
$⇒ α = \frac{2}{3}$
so,
$T = ( \frac{2}{3}, \frac{2}{3}, \frac{2}{3})$
$PT = \sqrt\frac{38}{3}$
$⇒ QT = \frac{4}{\sqrt3}$
So, Area
$= ( \frac{1}{2} × \frac{4}{\sqrt3} × \frac{\sqrt{38}}{\sqrt3} ) . 2$
$= \frac{4\sqrt{38}}{3} sq ∼ units$