Question

The line which contains all points $(x,y,z)$ which are of the form $\left(x,y,z\right)=\left(2,-2,5\right)+\lambda \left(1,-3,2\right)$ intersects the plane $2x-3y+4z=163$ at $P$ and intersects the $YZ$ plane at $Q$. If the distance $PQ$ is $a\sqrt{b}$ where $a,b\in N$ and $a>3$ then $(a+b)$ equals

• A. $23$
• B. $95$
• C. $27$
• D. none

Answer 1

The correct option is A $23$

Equation :$\frac{x-2}{1}=\frac{y+2}{-3}=\frac{z-5}{2}$

Any prove on this line $\left(r+2,-3r-2,2r+5\right)$

Line intersection prove that with plane $2x-3y+4z=163$

$2\left(r+2\right)-3\left(-3r-2\right)+4\left(2r+5\right)=163$

$2r+9r+8r+4+6+20=163$

$19r=133$

$r=7$

$P(9,-23,19)$

Prove that intersection of this line on $YZ$ plane $(x=0)$

$r+2=0$

$r=-2$

$Q=(0,4,1)$

$\begin{array}{c}PQ=\sqrt{81+{\left(-27\right)}^2+{\left(18\right)}^2}\\ =\sqrt{81+324+729}\\ =\sqrt{1134}\\ =\sqrt{9\times 126}\\ =\sqrt{9\times 6\times 21}\\ =\sqrt{3\times 3\times 3\times 2\times 3\times 7}\\ =3^2\sqrt{14}\\ =9\sqrt{14}\\ \therefore a=9,b=14\\ a+b=23\end{array}$

Then,

Option $A$ is correct answer.l