Let p1: x-2 y+3 z=5 and p2: 2 x+3 y+z+4=0 be two planes. If P is the foot of the perpendicular dropped from the origin ' O ' to the line of intersection of the planes, then

Question

Let p1: x-2 y+3 z=5 and p2: 2 x+3 y+z+4=0 be two planes. If P is the foot of the perpendicular dropped from the origin ' O ' to the line of intersection of the planes, then (A) vectors O P, hati-2 hatj+3 hatk and 2 hati+3 hatj+ hatk are coplanar. (B) If Q is another point on the line of intersection of planes then equation of the plane containing a triangle O P Q is 14 x +7 y +17 z =0. (C) Equation of the plane perpendicular to the line of intersection of the planes and passing through (1,1,1) is 11 x -5 y -7 z +1=0. (D) If N 1 N 2 are foot of the perpendicular from the origin ' O ' to the planes p 1 P 2 then ON 1+ ON 2 is equal to (9/√14).

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/abc5e65edc8cc6b68e9faf6dad4a653f-.png" /> (A) Obviously true. (B) p 1+λ p 2=0 which passes through (0,0,0) (C) Vector along the line of intersection of planes is | hat i hat j hat k 1 -2 3 2 3 1|=-11 hat i +5 hat j +7 hat k ∴-11(x-1)+5(y-1)+7(z-1)=0 11 x-5 y-7 z+1=0 (D) ON 1+ ON 2=(5/√14)+(4/√14)=(9/√14)

Q. Let $p_{1} : x - 2 y + 3 z = 5$ and $p_{2} : 2 x + 3 y + z + 4 = 0$ be two planes. If $P$ is the foot of the perpendicular dropped from the origin ' $O$ ' to the line of intersection of the planes, then

vectors $OP, \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $2 \hat{i} + 3 \hat{j} + \hat{k}$ are coplanar.

If $Q$ is another point on the line of intersection of planes then equation of the plane containing a triangle $OPQ$ is $14 x + 7 y + 17 z = 0$ .

Equation of the plane perpendicular to the line of intersection of the planes and passing through $(1, 1, 1)$ is $11 x - 5 y - 7 z + 1 = 0$ .

If $N_{1} & N_{2}$ are foot of the perpendicular from the origin ' $O$ ' to the planes $p_{1} & P_{2}$ then $O N_{1} + O N_{2}$ is equal to $\frac{9}{14}$ .

Q. Let p1​:x−2y+3z=5 and p2​:2x+3y+z+4=0 be two planes. If P is the foot of the perpendicular dropped from the origin ' O ' to the line of intersection of the planes, then

vectors OP,i^−2j^​+3k^ and 2i^+3j^​+k^ are coplanar.

If Q is another point on the line of intersection of planes then equation of the plane containing a triangle OPQ is 14x+7y+17z=0.

Equation of the plane perpendicular to the line of intersection of the planes and passing through (1,1,1) is 11x−5y−7z+1=0.

If N1​&N2​ are foot of the perpendicular from the origin ' O ' to the planes p1​&P2​ then ON1​+ON2​ is equal to 14​9​.

(A) Obviously true. (B) p1​+λp2​=0 which passes through (0,0,0) (C) Vector along the line of intersection of planes is ∣∣​i^12​j^​−23​k^31​∣∣​=−11i^+5j^​+7k^ ∴−11(x−1)+5(y−1)+7(z−1)=0 11x−5y−7z+1=0 (D) ON1​+ON2​=14​5​+14​4​=14​9​

Q. Let $p_{1} : x - 2 y + 3 z = 5$ and $p_{2} : 2 x + 3 y + z + 4 = 0$ be two planes. If $P$ is the foot of the perpendicular dropped from the origin ' $O$ ' to the line of intersection of the planes, then

vectors $OP, \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $2 \hat{i} + 3 \hat{j} + \hat{k}$ are coplanar.

If $Q$ is another point on the line of intersection of planes then equation of the plane containing a triangle $OPQ$ is $14 x + 7 y + 17 z = 0$ .

Equation of the plane perpendicular to the line of intersection of the planes and passing through $(1, 1, 1)$ is $11 x - 5 y - 7 z + 1 = 0$ .

If $N_{1} & N_{2}$ are foot of the perpendicular from the origin ' $O$ ' to the planes $p_{1} & P_{2}$ then $O N_{1} + O N_{2}$ is equal to $\frac{9}{14}$ .