Question:
Let $A$ be a point on the line
$\overrightarrow{\mathrm{r}}=(1-3 \mu) \hat{\mathrm{i}}+(\mu-1) \hat{\mathrm{j}}+(2+5 \mu) \hat{\mathrm{k}}$ and $\mathrm{B}(3,2,6)$
be a point in the space. Then the value of $\mu$ for which the vector $\overrightarrow{\mathrm{AB}}$ is parallel to the plane
$x-4 y+3 z=1$ is :
Correct Option: , 3
Solution:
Let point $\mathrm{A}$ is
$(1-3 \mu) \hat{\mathrm{i}}+(\mu-1) \hat{\mathrm{j}}+(2+5 \mu) \hat{\mathrm{k}}$
and point $B$ is $(3,2,6)$
then $\overrightarrow{\mathrm{AB}}=(2+3 \mu) \hat{\mathrm{i}}+(3-\mu) \hat{\mathrm{j}}+(4-5 \mu) \hat{\mathrm{k}}$
which is parallel to the plane $x-4 y+3 z=1$
$\therefore 2+3 \mu-12+4 \mu+12-15 \mu=0$
$8 \mu=2$
$\mu=\frac{1}{4}$