If the foot of the perpendicular drawn from (1,9,7) to the line passing through the point …

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 $x+2y+z=0$ and $3y-z=3$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+\gamma$ is equal to:

Accepted Answer

Direction of line

$$
\begin{aligned}
\vec{b} & =\left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 0 & 3 & -1 \end{array}\end{aligned}$$\right| \\
\& =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\
\& =-5 \hat{i}+\hat{j}+3 \hat{k}

\]

Equation of line

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

Let foot of perpendicular be $=(-5 k+3, k+2,3 k+1)$

$\Rightarrow(-5 k+2)(-5)+(k-7)(1)+(3 k-6)(3)=0$

Or $25 k-10+k-7+9 k-18=0$

Or $k=1$

$\alpha+\beta+\gamma=-k+6=5$

Solutions · 1