Let $L_1$ be the line of intersection of the planes given by the equations $2x + 3y + z = 4$ and $x + 2y + z = 5$. Let $L_2$ be the line passing through the point $P(2,-1, 3)$ and parallel to $L_1$. Let M denote the plane given by the equation $2x + y - 2z = 6$. Suppose that the line $L_2$ meets the plane M at the point Q. Let R be the foot of the perpendicular drawn from P to the plane M. Then which of the following statements is (are) TRUE?

Let L₁ be the line of intersection of the planes given by the equations 2x + 3y + z = 4 and x + 2y + z = 5. Let L₂ be the line passing through the point P(2,-1, 3) and parallel to L₁. Let M denote the plane given by the equation 2x + y - 2z = 6. Suppose that the line L₂ meets the plane M at the point Q. Let R be the foot of the perpendicular drawn from P to the plane M. Then which of the following statements is (are) TRUE?

$L_1:\ 2x+3y+z=4,\quad x+2y+z=5$

$L_2 \parallel L_1,\quad L_2 \ni P(2,-1,3)$

$M:\ 2x+y-2z=6$

$L_2 \cap M = Q$

$R = \text{foot of } \perp \text{ from } P \text{ to } M$

$\text{verify statements } (1)\text{–}(4)$

$1.\ \text{dir of } L_1 = \vec n_1 \times \vec n_2$

$2.\ L_2 = P + t\vec d \Rightarrow Q = L_2 \cap M$

$3.\ R = \text{foot from } P \text{ to } M;\ \text{verify } (1)\text{--}(4)$

$\vec n_1 = (2,3,1),\quad \vec n_2 = (1,2,1)$

d⃗ = n⃗₁ × n⃗₂ = î ĵ k̂ 2 3 1 1 2 1 = (1,-1,1)

$x + y = -1 \Rightarrow A(-1,0,6)$

$L_1:\ \vec r = (-1,0,6) + \lambda(1,-1,1)$

$L_2:\ \vec r = (2,-1,3) + t(1,-1,1)$

$M:\ 2x+y-2z=6$

$2(2+t)+(-1-t)-2(3+t)=6 \Rightarrow -3-t=6$

$t=-9 \Rightarrow Q(-7,8,-6)$

$d(P,M) = \dfrac{|2(2)+(-1)-2(3)-6|}{\sqrt{4+1+4}} = \dfrac{9}{3} = 3$

$R = (4,0,1),\quad \vec{PR} = R-P = (2,1,-2)$

$\vec{PQ} = Q-P = (-9,9,-9)$

$|PQ| = \sqrt{81+81+81} = 9\sqrt{3}\ \Rightarrow\ (1)\ \text{TRUE}$

$\vec{QR} = R-Q = (11,-8,7)$

$|QR| = \sqrt{121+64+49} = \sqrt{234} \ne 15\ \Rightarrow\ (2)\ \text{FALSE}$

$|PQ \times PR| = |(-9,-36,-27)| = 3\sqrt{234}$

[PQR] = 1/2|PQ× PR| = 3/2√(234) ⇒ (3) TRUE

$\cos\theta = \dfrac{\vec{PQ}\cdot\vec{PR}}{|PQ|\,|PR|} = \dfrac{9}{9\sqrt{3}\cdot 3} = \dfrac{1}{3\sqrt{3}}$

$\theta = \cos^{-1}\!\left(\dfrac{1}{3\sqrt{3}}\right) \ne \cos^{-1}\!\left(\dfrac{1}{2\sqrt{3}}\right) \Rightarrow (4)\ \text{FALSE}$

$\text{TRUE statements: } (1)\ \text{and}\ (3)$

$\boxed{(1)\ \text{and}\ (3)}$