Question

Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$ If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to_______

Answer 1

Correct Answer: 158

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 3& -2\\ 1& -1& 2\end{array}\right|=4\hat{i}-4\hat{j}-4\hat{k}$
$\therefore$ Equation of line is $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-1}{-1}$
Let $Q$ be $(5,3,8)$ and foot of $\perp$ from $Q$ on this line be $R$.
Now, $R\equiv (k+2,-k+3,-k+1)$
$DR$ of $QR$ are $(k-3,-k,-k-7)$
$\therefore (1)(k-3)+(-1)(-k)+(-1)(-k-7)=0$
$\Rightarrow k=-\frac{4}{3}$
$\therefore {\alpha }^2={\left(\frac{13}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2+{\left(\frac{17}{3}\right)}^2=\frac{474}{9}$
$\therefore 3{\alpha }^2=158$
So, the correct answer is 158.

Answer 2

Let the direction ratios of the given line be: \[ \left| \hat{i} \, \hat{j} \, \hat{k} \right| \left| 1 \, 3 \, -2 \right| \left| 1 \, -1 \, 2 \right| = 4\hat{i} - 4\hat{j} - 4\hat{k} \] Thus, the equation of the line is: \[ \frac{x - 2}{1} = \frac{y - 3}{-1} = \frac{z - 1}{2} \] Let \( Q = (5, 3, 8) \), and the foot of the perpendicular from \( Q \) on this line be \( R \). The direction ratios of QR are: \[ \text{DR of QR} = (-3, -3, -7) \] Thus, the distance \( \alpha^2 \) is: \[ \alpha^2 = \text{Distance} = 158 \quad \Rightarrow \quad 3\alpha^2 = 158 \]