The equations of the coordinate planes is given below:

yz:x=0, zx:y=0, xy:z=0

Consider in xy plane z=0.

Substitute 0 for z in the equation of line as given below,

\(\displaystyle{z}={0}\Rightarrow-{3}+{t}={0}\)

t=3, now x=3+2t

At t=3,

x=3+2*3=3+6=9

At t=3, y=5+7*3=5+21=26

Point is (9,26,0)

Consider in yz plane x=0.

Substitute 0 for x in the equation of line as given below,

\(\displaystyle{x}={0}\Rightarrow{3}+{2}{t}={0}\)

2t=-3. now

At \(\displaystyle{t}=-\frac{{3}}{{2}}\)

\(\displaystyle{y}={5}+{7}\cdot{\left(-\frac{{3}}{{2}}\right)}=\frac{{{10}-{21}}}{{2}}=-\frac{{11}}{{2}}\)

At \(\displaystyle{t}=-\frac{{3}}{{2}}\)

\(\displaystyle{z}=-{3}-\frac{{3}}{{2}}=\frac{{-{6}-{3}}}{{2}}=-\frac{{9}}{{2}}\)

Point is \(\displaystyle{\left({0},-\frac{{11}}{{2}},-\frac{{9}}{{2}}\right)}\)

Consider in zx plane y=0.

Substitute 0 for y in the equation of line as given below,

\(\displaystyle{y}={0}\Rightarrow{5}+{7}{t}={0}\)

7t=-5, now

At \(\displaystyle{t}=-\frac{{5}}{{7}}\)

\(\displaystyle{x}={3}+{2}\cdot{\left(-\frac{{5}}{{7}}\right)}=\frac{{{21}-{10}}}{{7}}=\frac{{11}}{{7}}\)

At t=-5/7ZSK

\(\displaystyle{z}=-{3}-\frac{{5}}{{7}}=-{21}-\frac{{5}}{{7}}=-\frac{{26}}{{7}}\)

Point is \(\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}\)

Thus, the point of intersections on the coordinate planes are as given below:

xy:(9,26,0)

yz:\(\displaystyle{\left({0},\frac{{11}}{{2}},\frac{{9}}{{2}}\right)}\)

zx:\(\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}\)

yz:x=0, zx:y=0, xy:z=0

Consider in xy plane z=0.

Substitute 0 for z in the equation of line as given below,

\(\displaystyle{z}={0}\Rightarrow-{3}+{t}={0}\)

t=3, now x=3+2t

At t=3,

x=3+2*3=3+6=9

At t=3, y=5+7*3=5+21=26

Point is (9,26,0)

Consider in yz plane x=0.

Substitute 0 for x in the equation of line as given below,

\(\displaystyle{x}={0}\Rightarrow{3}+{2}{t}={0}\)

2t=-3. now

At \(\displaystyle{t}=-\frac{{3}}{{2}}\)

\(\displaystyle{y}={5}+{7}\cdot{\left(-\frac{{3}}{{2}}\right)}=\frac{{{10}-{21}}}{{2}}=-\frac{{11}}{{2}}\)

At \(\displaystyle{t}=-\frac{{3}}{{2}}\)

\(\displaystyle{z}=-{3}-\frac{{3}}{{2}}=\frac{{-{6}-{3}}}{{2}}=-\frac{{9}}{{2}}\)

Point is \(\displaystyle{\left({0},-\frac{{11}}{{2}},-\frac{{9}}{{2}}\right)}\)

Consider in zx plane y=0.

Substitute 0 for y in the equation of line as given below,

\(\displaystyle{y}={0}\Rightarrow{5}+{7}{t}={0}\)

7t=-5, now

At \(\displaystyle{t}=-\frac{{5}}{{7}}\)

\(\displaystyle{x}={3}+{2}\cdot{\left(-\frac{{5}}{{7}}\right)}=\frac{{{21}-{10}}}{{7}}=\frac{{11}}{{7}}\)

At t=-5/7ZSK

\(\displaystyle{z}=-{3}-\frac{{5}}{{7}}=-{21}-\frac{{5}}{{7}}=-\frac{{26}}{{7}}\)

Point is \(\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}\)

Thus, the point of intersections on the coordinate planes are as given below:

xy:(9,26,0)

yz:\(\displaystyle{\left({0},\frac{{11}}{{2}},\frac{{9}}{{2}}\right)}\)

zx:\(\displaystyle{\left(\frac{{11}}{{7}},{0},-\frac{{26}}{{7}}\right)}\)