Question

If the foot of the perpendicular drawn from $(1,9,7)$ to the line passing through the point $(3,2,1)$ and parallel to the planes $x+2 y+z=0$ and $3 y-z=3$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

• A. $-1$
• B. 1
• C. 3
• D. 5

Answer 1

The Correct Option is D

The correct answer is (D) : 5

 

Direction ratio of line $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 0& 3& -1\end{array}\right|$
$=\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3)$
$=-5\hat{i}+\hat{j}+3\hat{k}$
$M(-5\lambda +3,\lambda +2,3\lambda +1)$
$\overrightarrow{\mathrm{PM}}\perp (-5\hat{i}+\hat{j}+3\hat{k})$
$-5(-5\lambda +2)+(\lambda -7)+3(3\lambda -6)=0$
$\Rightarrow 25\lambda +\lambda +9\lambda -10-7-18=0$
$\Rightarrow \lambda =1$
Point $M=(-2,3,4)=(\alpha ,\beta ,\gamma )$
$\alpha +\beta +\gamma =5$

Answer 2

Step 1: Determine the direction ratios of the line 

The given line passes through the point \( (3, 2, 1) \) and is parallel to the intersection of the planes:

\[ x + 2y + z = 0 \quad \text{(Equation 1)} \] \[ 3y - z = 3 \quad \text{(Equation 2)} \]

From Equation (2):

\[ z = 3y - 3 \]

Substitute this into Equation (1):

\[ x + 2y + (3y - 3) = 0 \] Simplify: \[ x + 5y - 3 = 0 \quad \Rightarrow \quad x = -5y + 3 \]

The direction ratios of the line are proportional to \( x = -5, y = 1, z = 3 \). Thus, the direction ratios of the line are \( (-5, 1, 3) \).

Step 2: Parametrize the line

The line passing through \( (3, 2, 1) \) with direction ratios \( (-5, 1, 3) \) is given by the parametric equations:

\[ x = 3 - 5t, \quad y = 2 + t, \quad z = 1 + 3t, \quad t \in \mathbb{R}. \] 

Step 3: Perpendicular from \( (1, 9, 7) \) to the line

Let the foot of the perpendicular from \( (1, 9, 7) \) to the line be \( (\alpha, \beta, \gamma) \). Substitute \( (\alpha, \beta, \gamma) = (3 - 5t, 2 + t, 1 + 3t) \).

The vector joining \( (1, 9, 7) \) to \( (\alpha, \beta, \gamma) \) is:

\[ \vec{V} = \left( (3 - 5t) - 1, (2 + t) - 9, (1 + 3t) - 7 \right) \] Simplify: \[ \vec{V} = (2 - 5t, -7 + t, -6 + 3t) \]

The vector \( \vec{V} \) is perpendicular to the direction vector \( (-5, 1, 3) \). Therefore, the dot product must be zero:

\[ \vec{V} \cdot (-5, 1, 3) = 0 \] Substitute \( \vec{V} \): \[ (2 - 5t)(-5) + (-7 + t)(1) + (-6 + 3t)(3) = 0 \] Simplify: \[ -10 + 25t - 7 + t - 18 + 9t = 0 \] Combine like terms: \[ 25t + t + 9t = 35 \quad \Rightarrow \quad 35t = 35 \quad \Rightarrow \quad t = 1. \] 

Step 4: Find \( (\alpha, \beta, \gamma) \)

Substitute \( t = 1 \) into the line equation:

\[ \alpha = 3 - 5(1) = -2, \quad \beta = 2 + 1 = 3, \quad \gamma = 1 + 3(1) = 4. \] 

Step 5: Calculate \( \alpha + \beta + \gamma \) \[ \alpha + \beta + \gamma = -2 + 3 + 4 = 5. \]